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PHILLIPS'LOOMIS  MATHEMATICAL   SERIES 


ELEMENTS    OF   TRIGONOMETRY 


PLANE   AND    SPHERICAL 


BY 
ANDREW    W.   PHILLIPS,   Ph.D. 

AND 

WENDELL   M.  STRONG,  Ph.D. 

YALE  UNIVERSIXy 


NEW  YORK  .:•  CINCINNATI  ■:•  CHICAGO 

AMERICAN     BOOK     COMPANY 


Copyright,  1898,  by  Harper  &  BROTHKRa. 

All  rights  reserved. 
W.  P.     18 


Xibrary 


3-51 


PREFACE 


In  this  work  the  trigonometric  functions  are  defined  as 
ratios,  but  their  representation  by  lines  is  also  introduced  at 
the  beginning,  because  certain  parts  of  the  subject  can  be 
treated  more  simply  by  the  line  method,  or  by  a  combination 
of  the  two  methods,  than  by  the  ratio  method  alone. 

Attention  is  called  to  the  following  features  of  the  book : 

The  simplicity  and  directness  of  the  treatment  of  both  the 
Plane  and  Spherical  Trigonometry. 

The  emphasis  given  to  the  formulas  essential  to  the  solu- 
tion of  triangles. 

The  large  number  of  exercises. 

The  graphical  representation  of  the  trigonometric,  inverse 
trigonometric,  and  hyperbolic  functions. 

The  use  of  photo-engravings  of  models  in  the  Spherical 
Trigonometry. 

The  recognition  of  the  rigorous  ideas  of  modern  mathe- 
matics in  dealing  with  the  fundamental  series  of  trigo- 
nometry. 

The  natural  treatment  of  the  complex  number  and  the 
hyperbolic  functions. 

The  graphical  solution  of  spherical  triangles. 

Our  grateful  acknowledgments  are  due  to  our  colleague, 
Professor  James  Pierpont,  for  valuable  suggestions  regard- 
ing the  construction  of  Chapter  VI. 

We  are  also  indebted  to  Dr.  George  T.  Sellew  for  making 
the  collection  of  miscellaneous  exercises. 

Andrew  W.  Phillips, 
Wendell  M.  Strong. 

Yale  University,  December,  i8gS. 

104S2.80 


TABLE    OF    CONTENTS 


PLANE   TRIGONOMETRY 
CHAPTER  I 

THE    TRIGONOMETRIC    FUNCTIONS 

PAGE 

Angles I 

Definitions  of  the  Trigonometric  Functions 4 

Signs  of  the  Trigonometric  Functions '    .     .  8 

Relations  of  the  Functions 10 

Functions  of  an  Acute  Angle  of  a  Right  Triangle 13 

Functions  of  Complementary  Angles 14 

Functions  of  0°,  90°,  180°,  270°,  360° 15 

Functions  of  the  Supplement  of  an  Angle 16 

Functions  of  45°,  30°,  60° 17 

Functions  of  (—;«:),  (i8o°—x),  (180°+^),  (360°— .r)   ....  18 

Functions  of  (90°— jv),  (90°  +  ^),  (270°— j),  (270°+ _)')      ...  20 

CHAPTER   II 

THE    RIGHT    TRIANGLE 

Solution  of  Right  Triangles 22 

Solution  of  Oblique  Triangles  by  the  Aid  of  Right  Triangles     .  28 

CHAPTER    III 

TRIGONOMETRIC    ANALYSIS 

Proof  of  Fundamental  Formulas  (i  i)-( 1 4) 32 

Tangent  of  the  Sum  and  Difference  of  Two  Angles      ....  36 

Functions  of  Twice  an  Angle 36 

Functions  of  Half  an  Angle 36 

Formulas  for  the  Sums  and  Differences  of  Functions   ....  37 
The  Inverse  Trigonometric  Functions 39 


Vi  TABLE  OF  CONTENTS 

CHAPTER  IV 

THE  OBLIQUE  TRIANGLE 

rAOB 

Derivation  of  Formulas 41 

Formulas  for  the  Area  of  a  Triangle 44 

The  Ambiguous  Case 45 

The  Solution  of  a  Triangle : 

(I.)  Given  a  Side  and  Two  Angles 46 

(2.)  Given  Two  Sides  and  the  Angle  Opposite  One  of  Them     .  46 

(3.)  Given  Two  Sides  and  the  Included  Angle 48 

(4.)  Given  the  Three  Sides 49 

Exercises 50 

CHAPTER  V 

CIRCULAR  MEASURE— GRAPHICAL  REPRESENTATION 

Circular  Measure 55 

Periodicity  of  the  Trigonometric  Functions 57 

Graphical  Representation 58 

CHAPTER  VI 

COMPUTATION  OF  LOGARITHMS  AND  OF  THE  TRIGONOMETRIC  FUNC- 
TIONS—DE  MOIVRE'S  THEOREM— hyperbolic   FUNCTIONS 

Fundamental  Series 63 

Computation  of  Logarithms 64 

Computation  of  Trigonometric  Functions 68 

De  Moivre's  Theorem , 70 

The  Roots  of  Unity 72 

The  Hyperbolic  Functions 73 

CHAPTER  VII 

MISCELLANEOUS  EXERCISES 

Relations  of  Functions 78 

Right  Triangles 80 

Isosceles  Triangles  and  Regular  Polygons 83 

Trigonometric  Identities  and  Equations 84 

Oblique  Triangles 88 


TABLE  OF  CONTENTS  vii 


SPHERICAL  TRIGONOMETRY 
CHAPTER  VIII 

RIGHT  AND  QUADRANTAL  TRIANGLES 

PAGB 

Derivation  of  Formulas  for  Right  Triangles 93 

Napier's  Rules 94 

Ambiguous  Case 97 

Quadrantal  Triangles 98 

CHAPTER  IX 

OBLIQUE-ANGLED  TRIANGLES 

Derivation  of  Formulas 100 

Formulas  for  Logarithmic  Computation loi 

The  Six  Cases  and  Examples     .............  104 

Ambiguous  Cases 106 

Area  of  the  Spherical  Triangle 108 

CHAPTER  X 

APPLICATIONS  TO  THE  CELESTIAL  AND  TERRESTRIAL  SPHERES 

Astronomical  Problems no 

Geographical  Problems 113 

CHAPTER  XI 

GRAPHICAL   SOLUTION  OF  A  SPHERICAL  TRIANGLE IIJ 

CHAPTER  XII 

RECAPITULATION   OF  FORMULAS II9 

APPENDIX 

RELATION   OF  THE  PLANE,  SPHERICAL,  AND  PSEUDO-SPHERICAL 

TRIGONOMETRIES I25 

ANSWERS  TO   EXERCISES 129 


PLANE  TRIGONOMETRY 

CHAPTER   I 
THE  TRIGONOMETRIC   FUNCTIONS 

ANGLES 

1,  In  Trigonometry  the  size  of  an  angle  is  measured  by 
the  amount  one  side  of  the  angle  has  revolved  from  the 
position  of  the  other  side  to  reach  its  final  position. 

Thus,  if  the  hand  of  a  clock  makes  one-fourth  of  a  rev- 
olution, the  angle  through  which  it  turns  is  one  right  angle ; 
if  it  makes  one-half  a  revolution,  the  angle  is  two  right  an- 
gles; if  one  revolution,  the  angle  is  four  right  angles;  if  one 
and  one-half  revolutions,  the  angle  is  six  right  angles,  etc. 


;:! 


-^B 


B 

FIG.  2 


The  amount  the  side  OB  has  rotated  from  OA  to  reach  its  final  position 
may  or  may  not  be  equal  to  the  inclination  of  the  lines.  In  Fig.  i  it  is  equal 
to  this  inclination  ;  in  Fig.  4  it  is  not. 

Two  angles  may  have  the   same  sides  and   yet  be  different.     In  Fig.  2 

I 


PLANE    TRIGONOMETRY 

and  Fig.  4  the  positions  of  the  sides  of  the  angles  are  the  same  ;  yet  in 
Fig.  2  the  angle  is  two  right  angles,  in  Fig.  4  it  is  six  right  angles.  The 
addition  of  any  number  of  complete  revolutions  to  an  angle  does  not  change 
the  position  of  its  sides. 

Question. — Through  how  many  right  angles  does  the  hour-hand 
of  a  clock  revolve  in  6\  hours.'  the  minute-hand  .'' 

Question. — If  the  fly-wheel  of  an  engine  makes  100  revolutions  per 
minute,  through  how  many  right  angles  does  it  revolve  in  i  second  } 


Initial  line 


\\    RIGHT   ANGLES 


Initial  line 


5}   RIGHT   ANGLES 


Def. — The  first  side  of  the  angle — that  is.  the  side  from 
which  the  revolution  is  measured — is  the  initial  line;  the 
second  side  is  the  terminal  line. 

Def. — If  the  direction  of  the  revolution  is  opposite  to  that 
of  the  hands  of  a  clock,  the  angle  is  positive;  if  the  same 
as  that  of  the  hands  of  a  clock,  the  angle  is  negative. 


Initial  line 


Initial  line 

POSITIVE    ANGLE 


NEGATIVE   ANGLE 


The  angles  we  have  employed  as  illustrations — those  described 

by  the  hands  of  a  clock — are  all  negative  angles. 

2,  Angles  are  usually  measured  in  degrees,  minutes,  and 
seconds.  A  degree  is  one-ninetieth  of  a  right  angle,  a  min- 
ute is  one-sixtieth  of  a  degree,  a  second  is  one-sixtieth  of  a 
minute. 


THE    TRIGONOMETRIC  FUNCTIONS 


The  symbols  indicating  degrees,  minutes,  and  seconds  are  °  '  "; 
thus,  twenty-six  degrees,  forty-three  minutes,  and  ten  seconds  is 
written  26°  43'  10". 

5.  The  plane  about  the  vertex  of  an  angle  is  divided  into 
four  quadrants,  as  shown  in  the  figure ;  the  first  quadrant 
begins  at  the  initial  line. 


II 


III 


IV 


A 
V\ 

T 

Initial  Line 

m: 

IV 

THE  FOUR  QUADRANTS 


ANGLE  IN  1ST  QUADRANT 


ANGLE  IN  2D  QUADRANT 


ANGLE  IN  3D  QUADRANT 


ANGLE  IN  4TH  QUADRANT 


An  angle  is  said  to  be  in  a  certain  quadrant  if  its  terminal 
line  is  in  that  quadrant. 

EXERCISES 

4.  (I.)  Express  2^  right  angles  in  degrees,  minutes,  and  seconds. 
In  what  quadrant  is  the  angle  ? 

(2.)  What  angle  less  than  360°  has  the  same  initial  and  terminal 
lines  as  an  angle  of  745°  ? 

(3.)  What  positive  angles  less  than  720°  have  the  same  sides  as  an 
angle  of  — 73"  ? 

(4.)  In  what  quadrant  is  an  angle  of  — 890°? 


4  PLANE    TRIGONOMETRY 

DEFINITIONS   OF  THE   TRIGONOMETRIC  FUNCTIONS 

5,  The  trigonometric  functions  are  numbers,  and  are  de- 
fined as  the  ratios  of  lines. 

Let  the  angle  AOP  be  so  placed  that  the  initial  line  is 
horizontal,  and  from  /*,  any  point  of  the  terminal  line,  draw 
PS  perpendicular  to  the  initial  line. 


S    A 


ANGLB  IN  THK  i  ST  QUADRANT 


ANOLB  IN  THE  aD  QUADRANT 


ANGLU  IN  THE  3D  QDADRAMT 


ANCLB  IN  THK  4TH  QUADKANT 


Denote  the  angle  A  OP  by  x. 
SP 


-=r-p  —  sine  of  x  (written  sin  x). 


OS 
OP 


cosine  of  x  (written  cos;r). 


THE    TRIGONOMETRIC  FUNCTIONS 


SP 


^r^  =  tangent  of  x  (written  tanx). 


-T— ,=  cotangent  of  x  (written  cot;ir). 
o/- 

•r^  =  secant  of  x  (written  sec;tr). 
-^p=  cosecant  of  x  (written  csc;r). 

To  the  above  may  be  added  the  versed  sine  (written  versin)  and  coversed 
sine  (written  coversin),  which  are  defined  as  follows  : 

versin  x  =  i  —  cos  x;  coversin  x  =  i  —  sin  as. 

The  values  of  the  sine,  cosine,  etc.,  do  not  depend  upon 

what  point  of  the  terminal  line  is  taken  as  P,  but  upon  the 

angle. 


s  s' 


S'6 


For  the  triangles  OSP  and  OS'P'  being  similar,  the  ratio  of  any 
two  sides  of  OS'P'  is  equal  to  the  ratio  of  the  corresponding  sides 
of  OSP. 

Def. — The  sine,  cosine,  tangent,  cotangent,  secant,  and 
cosecant  of  an  angle  are  the  trigonometric  functions 
of  the  angle,  and  depend  for  their  value  on  the  angle 
alone. 

Q»  A  line  may  by  its  length  and  direction  represent  a 
number ;  the  magnitude  of  the  number  is  expressed  by  the 
length  of  the  line  ;  the  number  is  positive  or  negative  ac- 
cording to  the  direction  of  the  line. 


6  PLANE    TRIGONOMETRY 

7.  In  §  5,  if  the  denominators  of  tlie  several  ratios  be 
taken  equal  to  unity,  the  trigonometric  functions  will  be  rep- 
resented by  lines. 

St     SP 
Thus,  sin;ir=-^,=  — =  SP—'Ci\Q  number  represented  by 

the  line,  that  is,  the  ratio  of  the  line  to  its  unit  of  length. 

Hence  SP  may  represent  the  sine  of  x. 

In  a  similar  manner  the  other  trigonometric  functions 
may  be  represented  by  lines. 

In  the  following  figures  a  circle  of  unit  radius  is  described 
about  the  vertex  O  of  the  angle  A  OP,  this  angle  being  de- 
noted by  X.     Then  from  §  5  it  follows  that 


)    Cot              B 

^^/^ 

^^\ 

'\ 

\ 

e»        ^— 

^ 

Cos    \ 

^ 

1    s        o 

^ 

FIG.  3 


THE    TRIGONOMETRIC  FUNCTIONS  7 

SP  represents  the  sine  of  x. 
OS  represents  the  cosine  of  x. 
A  T  represents  the  tangent  of  x. 
BC  represents  the  cotangent  of  x» 
6^7"  represents  the  §ecant  of  x. 
OC  represents  the  cosecant  of  x. 

For  the  sake  of  brevity,  the  lines  SP,  OS,  etc.,  of  the  preceding  figures  are 
often  spoken  of  as  the  sine,  cosine,  etc. 

Hence,  we  may  also  define  the  trigonometric  functions 
in  general  terms  as  follows: 

If  a  circle  of  unit  radius  is  described  about  the  vertex  of 
an  angle, 

(I.)  The  sine  of  the  angle  is  represented  by  the  perpendicular 
upon  the  initial  line  from  the  intersection  of  the  terminal  line  with 
the  circumference. 

(2.)  The  cosine  of  the  angle  is  represented  by  the  s^^ent  of  the 
initial  line  extending  from  the  vertex  to  the  sine. 

(3.)  The  tangent  of  the  angle  is  represented  by  a  line  tangent  to 
the  circle  at  the  beginning  of  the  first  quadrant,  and  extending  from 
the  point  of  tangency  to  the  terminal  line. 

(4.)  The  cotangent  of  the  angle  is  represented  by  a  line  tangent 
to  the  circle  at  the  beginning  of  the  second  quadrant,  and  extending 
from  the  point  of  tangency  to  the  terminal  line. 

(5.)  The  secant  of  the  angle  is  represented  by  the  segment  of  the 
terminal  line  extending  from  the  vertex  to  the  tangent. 

(6.)  The  cosecant  of  the  angle  is  represented  by  the  segment  of 
the  terminal  line  extending  from  the  vertex  to  the  cotangent. 

The  definitions  in  §  5  are  called  the  ratio  definitions  of  the  trigonometric 
functions,  and  those  in  §  7  the  litie  definitions.  The  introduction  of  two 
definitions  for  the  same  thing  should  not  embarrass  the  student.  We  have 
shown  that  they  are  equivalent.  In  some  cases  it  is  convenient  to  use  the 
first  definition,  and  in  other  cases  the  second,  as  the  student  will  observe 
in  the  course  of  this  study.  It  is  therefore  important  that  he  should  be- 
come familiar  with  the  use  of  both. 


8 


PLANE   TRIGONOMETRY 


SIGNS   OF   THE   TRIGONOMETRIC   FUNCTIONS 

S»  Lines  are  regarded  as  positive  or  negative  according 
to  their  directions.  Thus,  in  the  figures  of  §  5,  OS  xs,  posi- 
tive if  it  extends  to  the  right  of  O  along  the  initial  line, 
negative  if  it  extends  to  the  left ;  SP  h  positive  if  it  extends 
upward  from  OA,  negative  if  it  extends  doivnivard.  OP,  the 
terminal  line,  is  dXv^d.ys,  positive. 

The  above  determines,  from  §  5,  the  signs  of  the  trigono- 
metric functions,  since  it  shows  the  signs  of  the  two  terms 
of  each  ratio. 

By  the  line  definitions  the  signs  may  be  determined  di- 
rectly. The  sine  and  tangent  are  positive  if  measured  up- 
ward from  OAy  and  fiegative  if  measured  downward. 

The  cosine  and  cotangent  are  positive  if  measured  to  the 
right  from  OB,  and  negative  if  measured  to  the  left. 

B         Cot-t-  Cot-     B 


p> 

Cof-\ 

;^ 

1      ] 

5        0 

u 

na.  3 


na  4 


THE   TRIGONOMETRIC  FUNCTIONS 


The  secant  and  cosecant  are  positive  if  measured  in  the 
same  direction  as  the  terminal  line,  0P\  negative  if  measured 
in  the  opposite  direction. 

The  signs  of  the  functions  of  angles  in  the  different  quadrants  are  as  follows  : 


Quadrant 

I 

II 

Ill 

IV 

Sine  and  cosecant 

+ 

+ 

- 

- 

Cosine  and  secant 

+ 

- 

- 

+ 

Tangent  and  cotangent 

+ 

- 

+ 

- 

9,  It  is  evident  that  the  values  of  the  functions  of  an 
angle  depend  only  upon  the  position  of  the  sides  of  the 
angle.  If  two  angles  differ  by  360°,  or  any  multiple  of  360°, 
the  position  of  the  sides  is  the  same,  hence  the  values  of 
the  functions  are  the  same. 


O    Cot     B 


Thus  in  Fig.  i  the  angle  is  120°  in  Fig.  2  the  angle  is  840°  yet 
the  lines  which  represent  the  functions  are  the  same  for  both  angles. 

EXERCISE 
Determine,  by  drawing  the  necessary  figures,  the  sign  of  tan  rooo°; 
cos  810°;  sin  760°;  cot  —70°;  cos  —  550°;  tan  —560°;  sec  300°;  cot 
1560°;  sin  130°;  cos  260°;  tan  310°. 


lO 


PLANE   TRIGONOMETRY 


RELATIONS  OF  THE  FUNCTIONS 
10,  By  §  5,  whatever  may  be  the  length  of  OP,  we  have 


SP 


OS 


SP 


OS 


OP 


OP 


B"        Cot         C 


We  have,  then,  from  Figs.  2  and  3, 


SP     .  since 

OS  COS  a;' 

OS  _  _  cosag 

SP  sin  as  ' 


Multiplying  (l)  by  (2), 


or 


tana?  cota;=l, 

ta.nx  = ;    cot;r  = . 

cotjf  tanx 


Again,  from  Figs.  2  and  3, 

OP 


_„  =seca;= • 

OS  cos  05  ' 

OP  _  1 

•y/"  sin  as 


From  Figs.  2  and  3,  OS'  +  SP'  =  OP\ 


sin^a;  +  cos^a?  =  1, 




^.    y 

f" 

l/c 

\ 

X       w 

\ 

/<j;\Cos 

0 

u 

n&s 


or 

and  sin'jr=  I  —  cos^jr ;     cos' j:  =  i  —  sin' jr. 

Also,  OA'^AT'=Or,  and  OB'  +  BC'  =  OC\ 
or  1  +  tan' a?  =  sec' a? ; 

1  +  cot'x  =  csc'x. 


(0 

(«) 

(3) 


(4) 
(5) 

(6) 


(7) 
(8) 


THE   TRIGONOMETRIC  FUNCTIONS  il 

The  angle  x  has  been  taken  in  the  first  quadrant ;  the 
results  are,  however,  true  for  any  angle.  The  proof  is  the 
same  for  angles  in  other  quadrants,  except  that  SP  be- 
comes negative  in  the  third  and  fourth  quadrants,  and  OS 
in  the  second  and  third. 

EXERCISES 
11,  (I.)  Prove  cosjT  sec;r=  i. 

(2.)  Prove  sin^  csc^  =  l. 
(3.)  Prove  tan  x  cos  x  =  sin  x. 
(4.)  Prove  sin  jr  -v/i  —  cos"  ^  =  i  — -  cos'^r. 

(5.)  Prove  tan  x  +  cot  x  =  — 

'  sin;t:cosjr 

(6.)  Prove  sin*;r  —  cos*x  =1—2  cos' jr. 

(7.)  Prove =  sin;r. 

^  cotJT  sec  jr 

(8.)  Prove  tan  x  sin  jr  -f  cos;r  =  sec:r. 

1^,  The  formulas  (i)-(8)  of  §  10  are  algebraic  equations 
connecting  the  different  functions  of  the  same  angle.  If 
the  value  of  one  of  the  functions  of  an  angle  is  given,  we 
can  substitute  this  value  in  one  of  the  equations  and  solve 
to  find  another  of  the  functions.  Repeating  the  process,  we 
find  a  third  function,  etc. 

In  solving  equation  (6),  (7),  or  (8)  a  square  root  is  extracted ; 
unless  something  is  given  which  determines  whether  to  choose  the 
positive  or  negative  square  root,  we  get  two  values  for  some  of 
the  functions.  The  reason  for  this  is  that  there  are  two  angles. 
less  than  360°  for  which  a  function  has  a  given  value. 

EXERCISES 
13.  (I.)  Given  x  less  than  90°  and  sin^  =  ^;  find  all  the  other 
functions  of  x. 

Solution. — 

cos  x=  ±  v'  I  — i=  ±  iv'  3. 

Since  x  is  less  than  go'^,  we  know  that  cosjt  is  positive. 


12  PLANE  TRIGONOMETRY 

H^ncc  cos3;=+^"v/3; 

cota;  =  ij^  =  Vi; 

sec*  =  — ^  =  1^3; 

cscap  =  j  =  2. 
f 

(2.)  Given  tan^  =  "  i  ^iid  jc  in  quadrant  IV ;  find  sin  a:  and  zoi&x. 

Solution. — 

sin  a;  _     ^^ 

cos  X        ^ ' 
hence  3  sin  a;  =  —  cos  x, 

sin^x  +  cos^a;  =  i ; 
hence  10  sin^  x=  1 ; 

sin  X = -  V^=  - 1^  VIo ; 

cos  X  =  T^Vio. 

(3.)  Given  sin(— 30°)=— | ;  find  the  other  functions  of  —30**. 

(4.)  Given  x  in  quadrant  III  and  sin;c  =  — ^;  find  all  the  other 
functions  of  x. 

(5.)  Given  y  in  quadrant  IV  and  sin^  =  — |;  find  all  the  other 
functions  ofy. 

(6.)  Given  cos  60°  =  ^ ;  find  all  the  other  functions  of  60°. 

(7.)  Given  sin  0°  =  o ;  find  cos  0°  and  tan  0°. 

(8.)  Given  tan  2  =  f  and  z  in  quadrant  I ;  find  the  other  functions 
of  2. 

(9.)  Given  00145°=  ^  J  ^^^  ^^^  ^^^  other  functions  of  45°. 

(10.)  Given  ia.ny—^\^$  and  cosj^  negative;  find  all  the  other 
functions  of  y. 

(11.)  Given  cot  30°  =  V3  ;  find  the  other  functions  of  30°. 

(12.)  Given  2  sinA:=  i  —  cosj;  and  x  in  quadrant  II;  find  sin^; 
and  cos;c. 

(13.)  Given  tan  x  -\-  cotJf=  3  and  x  in  quadrant  I ;  find  sinjc. 


THE    TRIGONOMETRIC  FUNCTIONS 


13 


FUNCTIONS  OF  AN   ACUTE   ANGLE  OF  A  RIGHT  TRIANGLE 

14i,  The  functions  of  an  acute  angle  of  a  right  triangle 
can  be  expressed  as  ratios  of  the  sides  of  the  triangle. 


J< 


Remark. — Triangles  are  usually  lettered,  as  in  Fig.  2,  the  capital 
letters  denoting  the  angles,  the  corresponding  small  letters  the  sides 
opposite. 

In  the  right  triangle  ABC,  by  §  5, 

^     .      BC     a  ^ 

^      AC     h       .    ^ 
co8^  =-r^=-  =  sin^; 

AB     c  ' 

AC     b 

eotA=4S,=~=tanB. 
BC     a 

15,  From  §  14,  for  an  acute  angle  of  a  right  triangle,  we  have 
side  opposite  angle. 

hypotenuse 
side  adjacent  to  angle, 
hypotenuse 
_    side  opposite  angle    . 
side  adjacent  to  angle' 


sme: 


tangent : 


side  adiacent  to  angle 

cotangent  =  — r^ — -. f— . 

^  side  opposite  angle 


(9) 


14  PLANE  TRIGONOMETRY 


FUNCTIONS  OF   COMPLEMENTARY  ANGLES 

16.  From  §  14,  we  have 

sin  ^  =  cos  ^=cos(90°  — ^); 

cos  ^  =  sin  ^  =  sin  (90°—^); 
tan  A  =  QO\.  ^  =  cot(90°  — ^); 
cot^  =  tan^=tan(90°-^> 

Because  of  this  relation  the  sine  and  cosine  are  called  co-func- 
tions of  each  other,  and  the  tangent  and  cotangent  are  called  co- 
functions  of  each  other. 

The  results  of  this  article  may  be  stated  thus : 
A  fiinctioji  of  an  acute  angle  is  equal  to  the  co-function  of 
its  complementary  angle. 

The  values  of  the  functions  of  the  different  angles  are  given  in  "  Trigo- 
nometric Tables."  By  the  use  of  the  principle  just  proved,  each  function 
of  an  angle  between  45°  and  90°  can  be  found  as  a  function  of  an  angle  less 
than  45°.  Consequently,  the  tables  need  to  be  constructed  for  angles  up  to 
45°  only.  The  tables  are  so  arranged  that  a  number  in  them  can  be  read 
either  as  a  function  of  an  angle  less  than  45°  or  as  the  co- function  of  the 
complement  of  this  angle. 

EXERCISES 

17,  (i.)  Express  as  functions  of  an  angle  less  than  45** : 
sin  70° ;  cos  89°  30' ;  tan  63" ; 

cos  60° ;  cot  47° ;  sin  72°  39'. 

(2.)  cos  X  =  sin  2X  ;  find  x. 
(3.)  tan  X  =  cot  ;^x  ;  find  x. 
(4.)  sin  2X  =  cos  ^x  ;  find  x. 
(5.)  cot(30°  —  x)  =  tan(30°  +  ^x)  ;  find  x. 
(6.)  A,  B,  and   C  are  the  angles  of  a  triangle ;   prove  that 
cos^^=sini(^+  C). 
Hint.—  A+B  +  C^iSo". 


THE    TRIGONOMETRIC  FUNCTIONS 


15 


FUNCTIONS  OF'O*,  90°,  180**,  270°,  AND  360** 

IS,  As  the  angle  x  decreases  towards  0°  (Fig.  i),  sin  a:  de- 
creases and  cos;*:  increases.     When  OF  comes  into  coincidence 
with  OA,  .S/* becomes  o,  and  (^^  becomes  (3^(  =  i). 
Hence                         sin  0° = o.    cos  0° = i . 


PIO.  3 


nG.4 


As  the  angle  x  increases  towards  90°  (Fig.  2),  sin^  increases 
and  cos^  decreases.    When  C'/' comes  into  coincidence  with  OB, 
•S/* becomes  0B{—\)  and  OS  becomes  o. 
Hence  sin  90°  =  !,     cos  90°=©. 

As  the  angle  x  decreases  towards  0°  (Fig.  3),  tana;  decreases 
and  cot:JC  increases.     When  OP  comes  into  coincidence  with  OA, 
^7* becomes  o  and  BC  has  increased  without  limit. 
Hence  tan  0°  =  o,     cot  0°  =  00 . 

As  the  angle  x  increases  towards  90°  (Fig.  4),  tan  a:  increases 
and  cot;c  decreases.     When  OP  comes  into  coincidence  with  OB, 
AThzs  increased  without  limit,  and  BC—o. 
Hence  tan  90°  =  00,     cot  90°=©. 

Remark. — By  cot  0°=  00  we  mean  that  as  the  angle  approaches  indefinitely 
near  to  0°  its  cotangent  increases  so  as  to  become  greater  than  any  finite  quan- 
tity we  may  choose.  The  symbol  co  does  not  denote  a  definite  number,  but 
simply  that  the  number  is  indefinitely  great. 


i6 


PLANE    TRIGONOMETRY 


In  every  case  where  a  trigonometric  function  becomes  indefinitely 
great  it  is  in  a  positive  sense  if  the  anglg  approaches  the  limiting 
value  from  one  side,  in  a  negative  sense  if  the  angle  approaches  the 
limiting  value  from  the  other  side.  Thus  coto°=-}-oo  if  the  angle 
decreases  to  o°,  but  cot  o°=  — oo  if  the  angle  increases  from  a  nega- 
tive angle  to  o**.  We  shall  not  often  need  to  distinguish  between 
-j-oo  and  — 00,  and  shall  in  general  denote  either  by  the  symbol  w. 

By  a  similar  method  the  functions  of  i8o°,  270°,  and  360°  may  be 
deduced.    The  results  of  this  article  are  shown  in  the  following  table : 


Angle 

o» 

90° 

180° 

270° 

360° 

sin 

0 

I 

0 

—  I 

0 

cos 

I 

0 

—  I 

0 

I 

tan 

0 

CO 

0 

00 

0 

cot 

CO 

0 

00 

0 

00 

19,  It  may  now  be  stated  that,  as  an  angle  varies,  its  sine  and  cosine 
can  take  on  values  from  —  i  to  -\- 1  only,  its  tangent  and  cotangent  all 
values  from  — 00  to  -^<xi,  its  secant  and  cosecant  all  values  from  —  00 
/<?  -f  00 ,  except  those  between  —  7  and  -f-  /• 


FUNCTIONS  OF  THE  SUPPLEMENT  OF  AN  ANGLE 
jUO.  Suppose  the  triangle  OPS  (Fig.  i)  equal  to  the  tri- 
angle (9/"5'(Fig.  2),  then  SP=S'P'  and  OS=OS',  and  the 
angle  AOP'  (Fig.  2)  is  equal  to  the  supplement  of  AOP 
(Fig.  I).  Also,  in  the  triangle  AOP'  (Fig.  3),  angle  AOF 
=  angle^(9/"(Fig.  2). 


FIG.  3 


THE    TRIGONOMETRIC  FUNCTIONS 


17 


(lO) 


It  follows  from  §§  5  and  8  that 

sin  (1§0°—  «)  =  sfn  a;; 
COS  (180°  —x)=.  —  cos  05 ; 
tan  (180°  —  a?)  =  —  tan  x ; 
cot  (180°—  a:)  =  —  cot  a;. 

The  results  of  this  article  may  be  stated  thus :  — 

The  sine  of  an  angle  is  equal  to  the  sine  of  its  supplement^ 

and  the  cosine,  tangent,  and  cotangent  are  each  equal  to  minus 

the  same  functions  of  its  supplement. 

The  principle  just  proved  is  of  great  importance  in  the  solution  of  tri- 
angles which  contain  an  obtuse  angle. 

FUNCTIONS  OF  45°,   30°,  AND  6o® 

21,  In  the  right  triangle  OSP  (Fig.  1)  angle  0  =  angle  P—\^* 
and  OP=i. 

Hence  OS=SP=^  ^^2. 

Therefore  sin45°  =  cos450=i-/2;  §814,16 

tan45°  =  cot45°=i. 

P 


iVT 


ivT 


In  equilateral  triangle  OP  A  (Fig.  2)  the  sides  are  of  unit  length, 
PS  bisects  angle  OP  A,  is  perpendicular  to  OA,  and  bisects  OA. 
Hence,  in  the  right  triangle  OPS,  0S=^,  SP=^y/i. 


Therefore  sin  30°  =  cos  60°  =  i ; 

cos  30°  =  sin  60°  =-^-\/3 ; 
tan  30°  =  cot  60°  =  i  ^/i ; 
cot  30°  =  tan  60°=  Vs- 


§14 


1 8  PLANE   TRIGONOMETRY 

22,  The  following  values  should  be  remembered : 


Angle 

OP 

300 

45° 

60° 

900 

sin 

o 

i 

*^/2 

i\/3 

I 

cos 

I 

iVi 

W2 

i 

0 

EXERCISES  r^ 

Prove  that  if  jr  =  30°, 

(I.)  sin  2jr=:2  sinx  cos^r; 

(2.)  cos  3x=:4  cos';ir  — 3  cos^;  , 

(3.)  cos  2x  =  cos*  X  —  sin*  jr ; 

(4.)  sin  yi:  =  3  sin;ir  cos*;r — sin*4r; 

2  tan  X 
(5.)  tan  2x  =  - — - — j— . 
^•"  I— tan*jr 

(6.)  Prove  that  the  equations  of  exercises  i  and  3  are  cor- 
rect if  x=  45° 

(7)  Prove  that  the  equations  of  exercises  (2)  and  (4)  are  cor- 
rect if  ;r=i2o°. 


The  following  three  articles,  §§  23-25,  are  inserted  for 
completeness.  They  include  the  functions  of  (90 — x)  and 
(180— ;t-),  which,  on  account  of  their  great  importance,  were 
treated  separately  in  §§  16  and  20. 


FUNCTIONS  OF  {—x),  {lSo°—x),  (l8o°  +  ;ir),  (360°  — 4r) 

23.  The  line  representing  any  function — as  sine,  cosine,  etc. 
— of  each  of  these  ano;les  has  the  same  length  as  the  line  repre- 
senting the  same  funciion  of  x. 

Thus  in  Figs.  2  and  3,  triangle  05'/''=triangle  OSP,  hence  SP=S  P', 
and  OS=OS'. 


THE   TRIGONOMETRIC  FUNCTIONS 
B  C        tf^  B 


19 


> 

/ 

> 

T 

L 

f 

^ 

\ 

A 

\ 

y 

/ 

0           S 

p 

/ 

\y^ 

-N 

T" 

7^ 

T 

1^ 

as 

< 

\ 

A 

I                ^ 

^ 

\ 

s 

/ 

v^ 

s 

i 

T* 

FIG.  3 


FIG.  4 


In  Figs.  I  and  4,  triangle  OSP' =triang\e  OSP.  hence  SP'=SP. 

In  Figs.  I,  2,  and  4,  triangle  OA  7"=triangle  OA  T,  hence  A  T'=A  T. 

In  Figs.  I,  2,  and  4,  triangle  6'^C'=triangle  OBC,  hence  BC'=BC. 

TTierefore  any  function  of  each  of  the  angles  (— ^),  (180°—*), 
(180°+^))  (360°— :c),  is  equal  in  numerical  value  to  the  same  function 
of  X.  Its  sign,  however,  depends  on  the  direction  of  the  line  repre- 
senting it. 

Putting  in  the  correct  sign,  we  obtain  the  following  table : 


sin  (—  jt)  =  —  sin  JT 
cos(— j;)  =  cosj: 
tan(—  jt)  =  —  tan  j: 
eot  (—  x)  =  —  cot  jc 

sin  (180°  +  jt)  =  —  sin  X 
cos  (i  80°  +  jt)  =  -  cos  A 
tan(l8o°-t-;r)  =  tan:t: 
oot(l8o°  +  x)  =  cotA: 


sin  (180°  —  x)=.  sin  jr 

cos  (180P  —  x)  =  —  cos  JT 

tan  (180°  —  x)  =  —  tan  jp 
cot  (180°  —  x)=  —  cotx 

sin  (360°  —  jr)  =  —  sin  x 
cos  (360°  —  jr)  =  cos  jr 
tan  (360°  —  x)=.—\Axix 
cot  (360°  —  jr)  =  —  cot* 


PLANE   TRIGONOMETRY 


FUNCTIONS   OF  (90°— jj/),  (90°+jJ/),  (270°-/),  (270° +jj/) 

24:,  The  line  representing  the  sine  of  each  of  these  angles  is 
of  the  same  length  as  the  line  representing  the  cosine  of  j;  the 
cosine,  tangent,  or  cotangent,  respectively,  are  of  the  same  length 
as  the  sine,  cotangent,  and  tangent  of  ^. 


B 

=/ 

/ 

T 

C 

f 

M 

\ 

T 
A 

1                 0 

c 

i'  s 

y 

/ 

Np; 

y^ 

Xys' 

\ 

1 

A 

\               ^ 

s 

— > 

y 
\ 

FIC 

i     4 

N 

t' 

For 

Triangle  OS' P'  =  triangle  OSP,  hence  S' P  =  OS,  and  05'  =  5/'. 
Triangle  OA  T  —  triangle  O^C,  hence  A  T'  =  BC. 
Triangle  OBC  =  triangle  OA  T,  hence  BC  —  AT. 

Therefore  any  function  of  each  of  the  angles  (90°—^),  (90°-|-_>'), 
(270°— ^y),  (270°+^),  is  equal  in  numerical  value  to  the  co-function 


THE    TRIGONOMETRIC  FUNCTIONS  i\ 

of  y.     Its  sign,  however,  depends  on  the  direction  of  the  line  repre- 
senting it. 

Putting  in  the  correct  sign,  we  obtain  the  following  table  : 
sin  (90°  ~  y)=.  cos^  sin  (go°  +  ^)  =  cos  j 

cos  (go° — y)z=  sin^  cos  (90°  +  jf )  =  —  sin>' 

tan  (90°  —  y)=-  cot_y  tan  (90°  +  /):=—  cot^* 

cot  (90°  —y)  =  \xa.y  cot  (90°  +  ;')=—  tan  jv 

sin  (270°  —y)=—  co&y  sin  (270°  +>')=—  cos^ 

cos  (270° —y)=—  sin^  cos  (270°  +  y)  =  siny 

tan  (270°  —y)  =  cot  j  tan  (270°  +  y)=  —  coty 

cot  (270°  —  y)  =  tany  cot  (270°  +y)=  ~  taay 

25,  Either  of  the  two  preceding  articles  enables  us  directly  to 
express  the  functions  of  any  angle,  positive  or  negative,  in  terms 
of  the  functions  of  a  positive  angle  less  than  90°. 

Thus,  sin  212°  =sin  (180°  +  32°)  =  —  sin 32°; 

cos  260°  =  cos  (270°—  loP)  =  —  sin  loP. 

EXERCISES 

(I.)  What  angles  less  than  360°  have  the  sine  equal  to  —  J -v/i  ?  the 
tangent  equal  to-v/3  ? 

(2.)  For  what  values  of  x  less  than  720°  is  sin^  =  -^-^2  ? 

(3.)  Find  the  sine  and  cosine  of  — 30°;  765°;  120°;  210°. 

(4.)  Find  the  functions  of  405°;  600°;  1125°;  —45°;  225°. 

(5.)  Find  the  functions  of  — 120°;  — 225°;  —420°;  3270°. 

(6.)  Express  as  functions  of  an  angle  less  than  45°  the  functions  of 
233°:  —197°;  894°. 

(7.)  Express  as  functions  of  an  angle  between  45°  and  90°,  sin  267°; 
tan  (  —  254°) ;  cos  950"^. 

(8.)  Given  cos  164°  =  —  .96,  find  sin  196° 

(9.)  Simplify  cos (90° +  ;r)  cos (270° — jr)  — sin(i8oo  — jr)sin(36o°  — ;r). 

,     .   „.       ,.,   sin(i8o°  — jr)        .     „        x  ,  i 

(10.)  Simplify -T-^ r (tan(9o°— .r)  + 


sln(27o°— .r)  sin' (270°  — jr) 

(II.)  Express  the  functions  of  {x  —  90°)  in  terms  of  functions  of  x. 


CHAPTER  II 


THE  RIGHT  TRIANGLE 

27*  To  solve  a  triangle  is  to  find  the  parts  not  given. 

A  triangle  can  be  solved  if  three  parts,  at  least  one  of 
which  is  a  side,  are  given.  A  right  triangle  has  one  angle, 
the  right  angle,  always  given  ;  hence  a  right  triangle  can 
be  solved  if  two  sides,  or  one  side  and  an  acute  angle,  are 
also  given. 

The  parts  of  the  right  triangle  not  given  are  found  by 
the  use  of  the  following  formulas: 
_  opposite  side 


(I)  sine 

(3)  tangent  = 


hypotenuse 
opposite  side 


adjacent  side 

(2)  cosine       =-r^ ; 

^  hypotenuse 


14 


adjacent  side 


(4)  cotangent  = 


adjacent  side 


opposite  side  ' 
(6)  B-igd'—A).  §16 

To  solve,  select  a  formula  in  which  two  given  parts  enter;  substituting 
in  this  the  given  values,  a  third  part  is  found.  Continue  this  method  till 
all  the  parts  are  found. 

In  a  given  problem  there  are  several  ways  of  solving  the  triangle ;  choose 
the  shortest. 

EXAMPLE 

The  hypotenuse  of  a  right  triangle  is  47.653,  a  side  is 

21.34;  find  the  remaining  parts  and  the  area. 

B 


THE   RIGHT   TRIANGLE 


23 


SOLUTION    WITHOUT   LOGARITHMS 

The  functions  of  angles  are  given 
in  the  table  of  "  Natural  Functions." 

•     A     '^      21.34 

sin/3  =-= — - — 

c     47-653 

47.653)21.3400(^4478 
190612 

227880 
190612 

372680 
333571 


391090 
381224 


9866 

sin  ^=.4478 
^=26°  36' 

6=c  cos  A 
=47.653  X.  8942 

47.653 
.8942 


95306 
190612 
428877 
381224 
42.6113126 
d=42.6if 

^=(90°- 26°  36) =63 

area=^a* 

=^X2i. 34x42. 61 

21.34 
42.61 

2134 
12804 
4268 
8536 
2)909.2974 
454.6487 
area=454.6 


24 


SOLUTION   EMPLOYING    LOGARITHMS 

It  is  usually  better  to  solve  triangles 
by  the  use  of  logarithms. 

The  logaritliiiis  of  the  functions  are 
given  in  the  tables  of  "  Logarithms  of 
Functions."* 

sm  A=  - 

c 

log  sin  ^ =log  a — log  c 

log  21.34  =1.32919 

log  47. 653  =1.67809 

sub. 

log  sin  7^=9.65  no— 10 

^=26°  36'  14" 


cos  A-=- 

c 

log  b = log  r + log  cos  A 
log  47. 653 =1.67809 
log  cos  26°  36'  14"  =9.95140— 10 
log  ^=1.62949 

^=42.608 


^=(90° -26°  36'  I4")=63''  23'  46" 

area  =  \ab 
log  area = log  ^+loga4-log^ 

log  i= 9- 69897  — 10 
log  21.34=1.32919 
log  42. 608  =  1 .  62949 
log  area^2. 65765 

area=454.62 


*  In  this  solution  the  five-place  table  of  the  "  Logarithms  of  Functions"  is 
used. 

f  No  more  decimal  places  are  retained,  because  the  figures  in  them  are  not 
accurate  ;  this  is  due  to  the  fact  that  the  table  of  "  Natural  Functions"  is  only 
four-place. 


24  PLANE    TRIGONOMETRY 


CHECK  ON   THE  CORRECTNESS   OF  THE  WORK 

=  90.263  X  5.043 
90.263 
5 -043 


270789 
361052 

a»  =  455. 196309 
Extracting    the    square    root,   a  = 
21.34,  which  proves  the  solution  cor- 
rect. 


=  90.261  X  5.045 


log  90.261  =  1.95550 
log   5.045  =  0.70286 
2)2.65836 
log2i.34=  1. 32918 
a  =  21.34,  which  proves  the  solu- 
tion correct. 


Remark. — The  results  obtained  in  the  solution  of  the  preceding 
exercise  without  logarithms  are  less  accurate  than  those  obtained  in 
:he  solution  by  the  use  of  logarithms ;  the  cause  of  this  is  that  four- 
place  tables  have  been  used  in  the  former  method,  five  place  in  the 
latter. 

EXERCISES 

28.  (I.)  In  a  right  triangle  b  =  96.42,  c  =  1 14.81 ;  find  a  and  A. 

(2.)  The  hypotenuse  of  a  right  triangle  is  28.453,  a  side  is  18.197; 
find  the  remaining  parts. 

(3.)  Given  the  hypotenuse  of  a  right  triangle  =  747.24,  an  acute 
angle  =23°  45';  find  the  remaining  parts. 

(4.)  Given  a  side  of  a  right  triangle  =  37.234,  the  angle  opposite 
=  54°  27' ;  find  the  remaining  parts  and  the  area. 

(5.)  Given  a  side  of  a  right  triangle  =  1. 1293,  the  angle  adjacent 
=  74°  13'  27";  find  the  remaining  parts  and  the  area. 

(6.)  In  a  right  triangle  A  =  i$°  22'  11",  ^  =  .01793;  find  d. 

(7.)  In  a  right  triangle  B  =  7i°  34'  53",  b  =  896.33  ;  find  a. 

(8.)  In  a  right  triangle  ^  =  3729.4,  ^  =  2869.1  ;  find  A. 

(9.)  In  a  right  triangle  a  =  1247,  l>=  1988  ;  find  c. 

(10.)  In  a  right  triangle  a  =  8.6432,  3  =  4.781$;  find  B. 

The  angle  of  elevation  or  depression  of  an  object  is  the 
angle  a  line  from  the  point  of  observation  to  the  object 
makes  with  the  horizontal. 


THE  RIGHT  TRIANGLE 
P 


25 


Thus  angle  x  (Fig.  i)  is  the  angle  of  elevation  oi  P  \i  O  is  the  point  of 
observation  ;  angle  y  (Fig.  2)  is  the  angle  of  depression  oi  P\i  O  is  the  point 
of  observation. 

(i  I.)  At  a  horizontal  distance  of  253  ft.  from  the  base  of  a  tower  the 
angle  of  elevation  of  the  top  is  60°  20';  find  the  height  of  the  tower. 

(12.)  From  the  top  of  a  vertical  clifif  85  ft.  high  the  angle  of  depres- 
sion of  a  buoy  is  24°3i'  22" ;  find  the  distance  of  the  buoy  from  the 
foot  of  the  cliff. 

(13.)  A  vertical  pole  31  ft.  high  casts  a  horizontal  shadow  45  ft. 
long;  find  the  angle  of  elevation  of  the  sun  above  the  horizon. 

(14.)  From  the  top  of  a  tower  115  ft.  high  the  angle  of  depression 
of  an  object  on  a  level  road  leading  away  from  the  tower  is  22°  13' 
44" ;  find  the  distance  of  the  object  from  the  top  of  the  tower. 

(15.)  A  rope  324  ft.  long  is  attached  to  the  top  of  a  building,  and 
the  inclination  of  the  rope  to  the  horizontal,  when  taut,  is  observed 
to  be  47°  2i'  17" ;  find  the  height  of  the  building. 

(16.)  A  light-house  is  150  ft.  high.  How  far  is  an  object  on  the 
surface  of  the  water  visible  from  the  top? 

[Take  the  radius  of  the  earth  as  3960  miles.] 

(17.)  Three  buoys  are  at  the  vertices  of  a  right  triangle ;  one  side 
of  the  triangle  is  17,894  ft.,  the  angle  adjacent  to  it  is  57°  23' 46". 
Find  the  length  of  a  course  around  the  three  buoys. 

(18.)  The  angle  of  elevation  of  the  top  of  a  tower  observed  from  a 
point  at  a  horizontal  distance  of  897.3  ft.  from  the  base  is  10°  27'  42"; 
find  the  height  of  the  tower. 

(19.)  A  ladder  42I-  ft.  long  leans  against  the  side  of  a  building;  its 
foot  is  25^  ft.  from  the  building.  What  angle  does  it  make  with  the 
ground  ? 

(20.)  Two  buildings  are  on  opposite  sides  of  a  street  120  ft.  broad 


26 


PLANE    TRIGONOMEIRY 


The  height  of  the  first  is  55  ft. ;  the  angle  of  elevation  of  the  top  of 
the  second,  observed  from  the  edge  of  the  roof  of  the  first,  is  26°  37'. 
Find  the  height  of  the  second  building. 

(21.)  A  mark  on  a  flag-pole  is  known  to  be  53  ft.  7  in.  above  the 
ground.  This  mark  is  observed  from  a  certain  point,  and  its  angle 
of  elevation  is  found  to  be  25°  34'.  The  angle  of  elevation  of  the  top 
of  the  pole  is  then  measured,  and  found  to  be  34°  17'.  Find  the 
height  of  the  pole. 

(22.)  The  equal  sides  of  an  isosceles  triangle  are  each  7  in.  long;  the 
base  is  9  in.  long.    Find  the  angles  of  the  triangle. 


Hint. — Draw  the  perpendicular  BD.  BD  bisects  the  base,  and  also  the 
angle  ABC. 

In  the  right  triangle  ABD,  AB—^  in.,  AD=i^\  in.,  hence  ABD  can 
be  solved. 

Angle  C=  angle  A,  angle  ABC=2  angle  A  BD. 

(23.)  Given  the  equal  sides  of  an  isosceles  triangle  each  13.44  in., 
and  the  equal  angles  are  each  63°  21' 42";  find  the  remaining  parts 
and  the  area.  '' 

(24.)  The  equal  sides  of  an  isosceles  triangle  are  each  377.22  in., 
the  angle  between  them  is  19°  55'  32".  Find  the  base  and  the  area 
of  the  triangle. 

(25.)  If  a  chord  of  a  circle  is  18  ft.  long,  and  it  subtends  at  the  centre 
an  angle  of  45°  31'  10" ,  find  the  radius  of  the  circle. 

(26.)  The  base  of  a  wedge  is  3.92  in.,  and  its  sides  are  each  13.25  in. 
long ;  find  the  angle  at  its  vertex. 


THE   RIGHT    TRIANGLE  Vj 

(27.)  The  angle  between  the  legs  of  a  pair  of  dividers  is  64®  45',  the 
legs  are  5  in.  long;  find  the  distance  between  the  points. 

(28.)  A  field  is  in  the  form  of  an  isosceles  triangle,  the  base  of  the 
triangle  is  1793.2  ft. ;  the  angles  adjacent  to  the  base  are  each  53"  27' 
49".     Find  the  area  of  the  field. 

(29.)  A  house  has  a  gable  roof.  The  width  of  the  house  is  30  ft., 
the  height  to  the  eaves  25^  ft.,  the  height  to  the  ridge-pole  33^  ft. 
Find  the  length  of  the  rafters  and  the  area  of  an  end  of  the  house. 

(30.)  The  length  of  one  side  of  a  regular  pentagon  is  29.25  in. ;  find 
the  radius,  the  apothem,  and  the  area  of  the  pentagon. 


Hint. — ^The  pentagon  is  divided  into  5  equal  isosceles  triangle^  by  Us  radii. 
Let  AOB  be  one  of  these  triangles.  AB—2g.2S  in.  ;  angle  AOB=^  of 
36o°=72°.  Find,  by  the  methods  previously  given,  OA,  OD,  and  the  area 
of  the  triangle  A  OB. 

These  are  the  radius  of  the  pentagon,  the  apothem  of  the  pentagon,  and 
\  the  area  of  the  pentagon  respectively. 

(31.)  The  apothem  of  a  regular  dodecagon  is  2;  find  the  perimeter. 

(32.)  A  tower  is  octagonal ;  the  perimeter  of  the  octagon  is  153.7  ft. 
Find  the  area  of  the  base  of  the  tower. 

(33.)  A  fence  extends  about  a  field  which  is  in  the  form  of  a  regular 
polygon  of  7  sides;  the  radius  of  the  polygon  is  6283.4  ft.  Find  the 
length  of  the  fence. 

(34.)  The  length  of  a  side  of  a  regular  hexagon  inscribed  in  a  circle 
is  3.27  ft. ;  find  the  perimeter  of  a  regular  decagon  inscribed  in  the 
same  circle. 

(35.)  The  area  of  a  field  in  the  form  of  a  regular  polygon  of  9  sides 
is  483930  sq.  ft. ;  find  the  length  of  the  fence  about  it. 


28 


PLANE   TRIGONOMETRY 


SOLUTION  OF  OBLIQUE  TRIANGLES  BY  THE  AID  OF 
RIGHT  TRIANGLES 
29*  Oblique  triangles  can  always  be  solved  by  the  aid  of 
right  triangles  without  the  use  of  special  formulas ;  the 
method  is  frequently,  however,  quite  awkward ;  hence,  in  a 
later  chapter,  formulas  are  deduced  which  render  the  solu- 
tion more  simple. 

The  following  exercises  illustrate  the  solution  by  means 
of  right  triangles : 

(I.)  In  an  oblique  triangle  a  =  3.72.  5  =  47*  52',  C=i09*  10';  find 
the  remaining  parts. 

The  given  parts  are  a  side  and  two  angles. 

C 


Draw  the  perpendicular  CD. 

.Solve  the  right  triangle  BCD. 

Having  thus  found  CD,  solve  the  right  triangle  A  CD. 

(2.)  In  an  oblique  triangle  a  =  89.7,  czz.  125.3,  ^=Z9°  8';  find  the 
remaining  parts. 

The  given  parts  are  two  sides  and  the  included  angle. 


D      c  =  I2.-).3 


THE  RIGHT   TRIANGLE 


29 


Hint. — Draw  the  perpendicular  CD. 

Solve  the  right  triangle  CBD. 

Having  thus  found  CD  and  AD{=c—DB),  solve  the  right  triangle  ACD. 

(3.)  In  an  oblique  triangle  01  =  3.67,^=5.81,  A  =  27°22,' ',  find  the 
remaining  parts. 

TAe  given  parts  are  two  sides  and  an  angle  opposite  one  of 

them. 

0 


Either  of  the  triangles  ACB,  ACBf  contains  the  given  parts,  and 
is  a  solution. 

There  are  two  solutions  when  the  side  opposite  the  given  angle  is 
less  than  the  other  given  side  and  greater  than  the  perpendicular, 
CD,  from  the  extremity  of  that  side  to  the  base.* 

/^i<«/.— Solve  the  right  triangle  ACD. 

Having  thus  found  CD,  solve  the  right  triangle  CDB  (or  CDB\ 

(4.)  The  sides  of  an  oblique  triangle  are  a=  34.3,  b a  5&6,  c s  55.12 ; 
find  the  angles. 

The  given  parts  are  the  three  sides. 


c=S&.12 


*  A  discussion  of  this  case  is  contained  in  a  later  chapter  on  the  solution 
of  oblique  triangles. 


30 


PLANE    TRIGONOMRCRY 


Hitii.'^ 

Let  DB=x, 

a;'-*«=CZ>'=^-(<r-*)». 

Hence 

o»=^-<»+2^. 

In  each  of  the  right  triangles  A  CD  and  BCD  the  hypotenuse  and  a  side 
are  now  known ;  hence  these  triangles  can  be  solved. 

(5.)  Two  trees,  A  and  B,  are  on  opjxjsite  sides  of  a  pond.  The 
distance  of  A  from  a  point  C  is  297.6  ft.,  the  distance  of  B  from  C  is 
864.4  ft.,  the  angle  ACB  is  87"  43'  12".     Find  the  distance  AB. 

(6.)  To  determine  the  distance  of  a  ship  A  from  a  point  B  on 
shore,  a  line,  BC,  800  ft.  long,  is  measured  on  shore ;  the  angles,  ABC 
and  ACB,  are  found  to  be  67°  43'  and  74°  21'  16"  respectively.  What 
is  the  distance  of  the  ship  from  the  point  B  ? 

(7.)  A  light-house  92  ft.  high  stands  on  top  of  a  hill ;  the  distance 
from  its  base  to  a  point  at  the  water's  edge  is  297.25  ft. ;  observed 
from  this  point  the  angle  of  elevation  of  the  top  is  46°  33'  15",  Find 
the  length  of  a  line  from  the  top  of  the  light-house  to  the  point. 

(8.)  The  sides  of  a  triangular  field  are  534  ft.,  679.47  ft.,  474.5  ft. 
What  are  the  angles  and  the  area  of  the  field  ? 

(9.)  A  certain  point  is  at  a  horizontal  distance  of  117!^  ft.  from  a 
river,  and  is  ii  ft.  above  the  river;  observed  from  this  point  the  angle 
ofdepression  of  the  farther  bank  isi"i2'.  Whatisthe  width  of  the  river? 

(lo.)  In  a  quadrilateral  A  BCD.  AB  =  1.41,  BC=  i  .05,  CD  =  1.76,  DA 
rsi.93,  angle  A=:7S°  21';  find  the  other  angles  of  the  quadrilateraL 


THE  RIGHT    TRIANGLE  3I 

Hint. — Draw  the  diagonal  DB. 

In  the  triangle  ABD  two  sides  and  an  included  angle  are  given,  hence  the 
triangle  can  be  solved. 

The  solution  of  triangle  ABD  gives  DB. 

Having  found  DB,  there  are  three  sides  of  the  triangle  Z'.fiC  known,  hence 
the  triangle  can  be  solved. 

(II.)  In  a  quadrilateral  ABCD,  AB=i2.i,  AD  =  9.7,  angle  A  — 
i7°  18',  angle  B  —  64°  49',  angle  Z>=  100°;  find  the  remaining  siaes. 

ffittt.— Solve  triangle  ABD  to  find  £D. 


CHAPTER   III 
TRIGONOMETRIC   ANALYSIS 

30,  In  this  chapter  we  shall  prove  the  following  funda- 
mental formulas,  and  shall  derive  other  important  formulas 
from  them : 

8lii(a;  +  y)  =  8lnaj  co§y +  cosic  siny,  (ii) 

§ln(aj  — j/)  =  §in£c  cosy  —  cosic  siny,  (12) 

€08(05  +  ?/)  =  cos  a;  cosy  — sin  a?  sini/,  (13) 

cos(a;  — y)  =  cos £C  cosy  +  sin ic  sin y;  (14) 

PROOF   OF   FORMULAS   (ll)-(l4) 

31,  Let  angle  AOQ=x,  angle  QOP=y\  then  angle  AOP 

The  angles  x  and  y  are  each  acute  and  positive,  and  in  Fig.  i 
(ar+j')  is  less  than  90°,  in  Fig.  2  {x-\-y')  is  greater  than  90°. 


In  both  figures  the  circle  is  a  anit  drde,  and  SP  is  perpendicular  to 
CA ;  hence  SP=%va.  {x  +/>  <?6'=s  co»  (*  +/> 


TRIGONOMETRIC  ANALYSIS  l\ 

Draw  DP  perpendicular  to  OQ ; 
then  Z>/*=sinj,     OD=.zo%y, 

angle  SPD-2.x\g\^  AOQ=^x. 

(Their  sides  being  perpendicular.) 
Draw  DE  perpendicular  to  OA,  DH  perpendicular  to  SP. 
Sin  {x  A-y)  =  SP^  ED  +  HP. 
ED={^\x\x)y.OD  =  syc\x  cosj. 

ED 
(For  OED  being  a  right  triangle,-— -=  sin  jr.) 

HP={cosx)  X  DP=  COS  X  sin  J. 

(For  HPD  being  a  right  triangle,  -— —  =  cosj:.) 

Therefore,  sin  (a; +  3/)  =  sin  x  cos  j/  + cos  as  gin  y.  (ll) 

Co%{x-^y)=OS=  OE-HD* 
OEz={cosx)x  OD  — cos X  cos/. 

OE 
(For  C^Z>  being  a  right  triangle,  -— —  =  cosx.) 

HD  =  {sin  x)  X  DP— sm  x  sxny. 

H  D 
(For  PHD  being  a  right  triangle,  -  -^  =  sinx.) 

Therefore,  cos  (05  + y)  =  cos  a;  cosy -sin  x  sin  j/.  (13) 

5^.  The  preceding  formulas  have  been  proved  only  for 
the  case  when  x  and  y  are  each  acute  and  positive.  The 
proof  can,  however,  readily  be  extended  to  include  all  values 
of  X  and  y. 

Let  y  be  acute,  and  let  x  be  an  angle  in  the  second  quad- 
lant;  then  ;r  =  (90°4-.r')  where  x'  is  acute, 
sin  {x  +_y)  =  sin  (90°  +  x'  +j) 

=  cos(;r'+j)  §24 

=  cos  x'  cosjj'— sin  .r'  sin^j' 

=  sin(90°  +  ;ir')  cosjj/  +  cos(90°  +  ;tr')sin_;'      §24 
=  sin-r  cosj-fcos;r  sinj. 

*  If  (.r  +  v)  is  greater  than  90°,  05  is  negative. 


34  PLANE   TRIGONOMETRY 

Thus  the  formula  has  been  extended  to  the  case  where 
one  of  the  angles  is  obtuse  and  less  than  i8o°.  In  a 
similar  way  the  formula  for  cos(,r4-j)  is  extended  to  this 
case. 

By  continuing  this  method  both  formulas  are  proved  to 
be  true  for  all  positive  values  of  x  and  y. 

Any  negative  angle  _;/ is  equal  to  a  positive  angle  y,  minus 
some  multiple  of  360°.  The  functions  of  y  are  equal  to 
those  of  y\  and  the  functions  of  {x-\-y)  are  equal  to  those 
of  (:«:+/).  §  9 

Therefore,  the  formulas  being  true  for  {x-\-y'),  are  true  for 

A  repetition  of  this  reasoning  shows  that  the  formulas  are 
true  when  both  angles,  x  and  y,  are  negative. 

33.  Substituting  the  angle  —y  for  y  in  formula  (n),  it 
becomes 

sin  (x—y)  =  sin  ;t:  cos  ( —y)  +  cos  x  sin  (  —_;/), 
But  cos(— y)  =  cosj,  and  sin  (— j)=  — sin_;'.      §23 

Therefore,  8in(ic  — 2/)  =  8lnic  cosy -cos  a?  sinj/.  (12) 

Substituting  {—y)  for  j  in  formula  (13),  it  becomes 
cos(;tr--jj')  =  cos;r  cos(— j)  — sin;r  sin  (—_;/), 
r=cos;r  cos^  +  sin;ir  sin^. 
Therefore,  co8{ic  —  y)  =  cos x  cos  y  +  sin  x  sin  y.*  (14) 

EXERCISES 

34*  (I.)  Prove  geometrically  where  x  and  y  are  acute  and  positive  : 
sin  {x  —j)  =  sin  ;r  cos/  —  cos  x  sin_y, 
cos(-r — _y)  =  cos;r  cos/  +  sin  jt  sin/. 

*  Formulas  (12)  and  (14)  are  proved  geometrically  in  §  34.  The  geometric 
proof  is  complicated  by  the  fact  that  OD  and  DF  are  functions  of  —y,  while 
the  functions  of  /  are  what  we  use. 


TRIGONOMETRIC  ANALYSIS 


35 


Hint.— Ang]e  AOQ—x,  angle  POQ=y,  and  angle  AOP=(x—y). 

Draw  PD  perpendicular  to  OQ. 
ThenZ)/'=sin(— 7)=— sinv;  but  DP  is  negative,  therefore  PD  taken 
as  positive  is  equal  to  sin  y : 

C?Z> = cos  ( —  j) = cos  ^'. 

Angle  IIPD=3iX\^e  AOQ=x.  their  sides  being  perpendicular. 

Draw  Z>iy  perpendicular  lO  SP,  DE  perpendicular  to  OA, 

^\n{x-y)=SP=ED-PH. 

From  right  triangle  OED,    ED=i{s\nx)  x  OD=s,\nx  cosy. 

From  right  triangle  DIIP,    PJ/=(cosx)  x  PD=cosx  siny. 

Therefore,  sin {x—y)=smx  cosj  — cosjt  siny. 

Cos  {x  -y)  =  0  S=  OE + DH. 

From  right  triangle  OE.D,     OE—{co%x)  X  OD-=co%x  cos_y. 

From  right  triangle  DIIP,     DH={smx)  x  PD  =  sinx  sin^. 

Therefore,  cos (jr —;!■)= cos ;>:  cosj  +  sinx  sin_;'. 

(2.)  Find  the  sine  and  cosine  of  (45°+.i-),  (30°— ;r),  (60°+ jr),  in  terms 
of  sin;r  and  cos;r. 

(3.)  Given  sin;ir=f,  sin_)'  =  -^,  x  and  j  acute;  find  sin(x-\-y)  and 
sin(;r — y). 

(4.)  Find  the  sine  and  cosine  of  75°  from  the  functions  of  30° and  45°. 
Hint.—  75°=(45°  +  30'=). 

(5.)  Find  the  sine  and  cosine  of  15°  from  the  functions  of  30° and  45°. 

(6.)  Given  x  and  y,  each  in  the  second  quadrant,  sin  x  =  ^,  sin_y  =  J ; 
find  sin(;r+j/)  and  cos  (;r —_>/). 

(7.)  By  means  of  the  above  formulas  express  the  sine  and  cosine  of 
(i8o°  — ;r),  (i8o°-|-;ir),  {270°— x),  (270°+ jr),  in  terms  of  sin^  and  cos;ir. 

(8.)  Prove  sin  (6o°-f  45")-!- cos  (60° +  45°)  =  00345°. 

(9.)  Given  sin45°  =  -|^-v/2,  cos45°==^-\/2  ;  find  sin  90°  and  cos  900. 

(10.)  Prove  that  sin(6o°+ ;r)  — sin(6o°  — -r)  =  sin-r. 


36  PLANE  TRIGONOMETRY 

TANGENT  OF  THE  SUM  AND  DIFFERENCE  OF  TWO  ANGLES 

55.  Tan(;ir-f-j/)  =  s^"(^+-^)  =  sin^  cosj  +  cos.r  sinj/  ^ 

cos(;ir4-^)      cos;ir  cos_>/— sin;trsin^ 

Dividing  each  term  of  both  numerator  and  denominator 

of  the  right-hand  side  of  this   equation  by  cos;ir  cos/,  and 

Sin 
remembering  that  —  =  tan,  we  have 
cos 

.      ,  ^       tan£c  +  tany  ,     . 

tan(a?  +  y)  = ^-.  (15) 

1  —  tan  X.  tan  y 

In  a  similar  way,  dividing  formula  (12)  by  formula  (14),  we 

obtain 

.       ,  .       tan oc  —  tan y  ,    ,. 

tan(ic-2/)  =  — — ^-.  (16) 

1  +  tana?  tan 2/ 

FUNCTIONS  OF  TWICE  AN  ANGLE 

56.  An  important  special  case  of  formulas  (11),  (13),  and 
(15)  is  when  y=x\  we  then  obtain  the  functions  of  2x  in 
terms  of  the  functions  of  x. 

From  (it),  sin(,r+;r)=sin^cos,t'-f  cos;rsin^. 
Hence  sin2ic=2siniccosic.  (17) 

From  (13),        cos2a!;  =  cos^a?  — sin'^ic.  (18) 

Since        cos^;r=  i  — sin^:r,  and  sin^;ir=  i— cos^;ir, 

we  derive  from  equation  (18), 

cos2;t:=  I  —  2sin^;ir,  (19) 

and  cos2;r=2  cos^;r— I.  (20) 

T-  /    -\         u.       «  2  tan  a?  z      . 

From  (15),       tan2x-=- — - — — .  (21) 

1  —  tan'^ic 

FUNCTIONS   OF  HALF  AN  ANGLE 

57.  Equations  ( 19)  and  (20)  are  true  for  any  angle  ;  there- 
fore for  the  angle  \x. 

From  ( 19),  cosx=  i  —  2  sin^  )yX  • 


or  sm*;tr  = 


TRIGONOMETRIC  ANALYSIS  37 

I  —  cos;tr 


2 


therefore,  8lii-Ja5=r  ±\/ •  (22) 

From  (20),  cos;tr  =  2  cos'^;ir  — I  : 

■  -         I  +  cos;ir 
or  cos^;i;  = -; 


therefore,  cos^=±\/ ^ *  (23) 

Dividing  (22)  by  (23),  we  obtain 

tan  ^  =  ±  \  / ,  ^??^ .  (24) 

FORMULAS   FOR   SUMS  AND   DIFFERENCES  OF  FUNCTIONS 
38,  From  formulas  (ii)-(i4),  we  obtain 

sin(;jr+7)  +  sin  (;ir  — jj')=:2sinjtr  cosj; 
sin  (;ir + jj/) — sin  (;ir — jj/) = 2  cos;»r  sinj^ ; 
cos  {x  +j/)  4-  cos  {x  —y) = 2  cos;r  cosjf ; 
cos  {x  +  j) — cos  {x  —y)  =  -  2  sin  ;r '  sin^. 
Let  u  =  {x-{-y)  a.nd  v  =  (x—y); 

then  x=^{u-\-v),  y=^{u^v). 

Substituting  in  the  above  equations,  we  obtain 

slnu  +  sinv  =  2  ain^iu  +  v) cos^(u  —  v)',  (25) 

sinu  —  sin  v  =  2  co§^(w  +  v)  sln^{u  —  v) ;  (26) 

cos  u  +  COS  V—  2  cos|-  (11+ v)  cos^  {u  —  v) ;  {2y) 

cosw-co8v=-28in^(t*  + v)  sln^{u  —  v),         (28) 
Dividing  (25)  by  (26), 

8in«^+8in7;     tan^(t*+v) 


sin  «* — sin  V     tan^(«* — v) 


(29) 


EXERCISES 

39.  Express  in  terms  of  functions  of  x,  by  means  of  the  formulas 
of  this  chapter, 


38  PLANE   TRIGONOMETRY 

(I.)  Tan(i8o°  — jr);  tan(i8oo  +  4:). 

(2.)  The  functions  of  {x  —  i8o°). 

(3.)  Sin(-r  —  90°)  and  003(^  —  90°). 

(4.)  Sin(^  — 270°),  and  005(^  —  270°). 

(5.)  The  sine  and  cosine  of  (45°— a:);  of  (45®+Jf). 

(6.)  Given  tan  45°=  i,  tan  io^  —  \  -y/J;  find  tan  75°;  tan  15°. 

,     X    T^  ^   ,  ^         COt.X  COlW-1  ,       ^ 

(7.)  Provecot  (ic+w)  = ; — ; V — •  (30) 

Hint. — Divide  formula  (13)  by  formula  (il). 

„     ^  cotic  cot  1/ 4-1  ,    , 

(8.)  Prove  cot [oc-y)  = ; ^ —  .  (31) 

COt*/-COtiC  *^ 

(9.)  Prove  cos  (30  -f- j)  —  cos  (30°  —y)  =  —  sin_y. 

(10.)  Prove  sin3jr  =  3  sin;r — 4  sin'jr. 

f/tnt. — Sin  3jr=sin  (x+2x). 

(I  I.)  Prove  cos  3^:  =  4  cos*  x  —  3  cos  x. 

(12.)  If  X  and  y  are  acute   and   tan;r  =  ^,  tan/  =  J,  prove  that 

(x-\-y)-4S°- 

^  ,  ,     ,       ^       i-f-tanjr 

(13.)  Prove  that  tan  (x  +  45°)  = • 

•^  I  —  tan  X 

(14.)  Given  sinj  =  |  and /acute;  find  sin  ^y,  cos^y,  and  tan  ^7. 

(15.)  Given   cos;r=— I  and  x  in  quadrant  II;    find    sin  2-r  anc 

cos  2X. 

(16.)  Given  cos  45°  =  ^  -\/2  ;  find  the  functions  of  22^°. 
(17.)  Given  tan  x  =  2  and  x  acute  ;  find  tan  ^x 
(18.)  Given  cos  30°  =  ^ -v/3  ;  find  the  functions  of  15°. 
(19.)  Given  cos9o°  =  o;  find  the  functions  of  45°. 
(20.)  Find  sin^jf  in  terms  of  sin  jr. 
(21.)  Find  cos5;r  in  terms  of  cos.r. 

(22.)  Prove  sin  (x  -\-y  -{-z)=:s\n  x  cosy  cos  ^•-f-cos  x  s'lny  cos  5'+cos  Jt 
cosy  sinz  —  sin  jr  s'lny  s'mz. 

Hint. — Sin(jr+j+2)=sin(x+^)  cos0+cos(jr+>')  sins. 
(23.)  Given  tan  2.r=:3  tan;i:;  find  x. 
(24.)  Prove  sin  32°  +  sin  28°  =  cos  2°, 
(25.)  Prove  tan  jr+cot^  =  2  csc2jr. 
(26.')  Prove  (sin  \x  +  cos  \x'^  =  i  4-  sin  x, 
(27.)   Prnvf  (sin  \.x      COS  ^r)-' =  r       sin  ;jr. 


TRIGONOMETRIC  ANALYSIS  -     39 

(28.)  Prove  cos  2X  =  cos^;^::  —  sin*,^. 

(29.)  Prove  tan  (45°  +  x)  +  tan  (45°  —  x)  =  2  sec  zx. 

,      .   _,  .  2  tanx 

(30.)  Prove  sin  2X  = 


(31.)  Prove  cos  2 jc  = 


I  +  tan^jf 
I  —  tan^.y 
I  +  tan^^c 


/      V   -r,         I  +  sin  2^     /tan  x  +  i  V 

(32.)  Prove — ' =   • 

I  —  sin  2JC     \tan x—  \J 

(^11)  Prove  tani^  = 


(34.)  Prove  cot  1^^  = 


I  +  cos  X 
sin;c 

I  —  COSJC 


(35.)  Express  as  a  product -•* 

^  cos^f  +  cos  J 

jj.. COS  a:— cosy  _  —  2  sin^(a:+>')  sinK^— y) 

cosac+cosy       2  COS  5  U+y)  COS  J  (a;— y) 

=  — tanKar+y)  tan|(ar— y). 

/..„  j^tan:r  +  tan  y 

(xdA  Express  as  a  product =^« 

^•^     "^         ^  COtJC  +  COtj 

/      \  T)  i.         4.  cos(;t:  +  y) 

(37.)  Prove  I  —  tan  ;c  tanj' =  — ^^ — -=^- 

COS  X  cos_y 

THE   INVERSE  TRIGONOMETRIC   FUNCTIONS 

40.  i?^. — The  expressions  sin"^«,  cos~'a,  tan~'«,  etc.,  de- 
note respectively  an  angle  whose  sine  is  a,  an  angle  whose 
cosine  is  a,  an  angle  whose  tangent  is  a,  etc.  They  are 
called  the  inverse  sine  of  a,  the  inverse  cosine  of  a,  the 
inverse  tangent  of  «,  etc.,  and  are  the  inverse  trigono- 
metric functions. 

Sin~^a  is  an  angle  whose  sine  is  equal  to  a,  and  hence  de- 
notes, not  a  single  definite  angle,  but  each  and  every  angle 
whose  sine  is  a. 

*  Since  quantities  cannot  be  added  or  subtracted  by  the  ordinary  operations 
with  logarithms,  an  expression  must  be  reduced  to  a  form  in  which  no  addition 
or  subtraction  is  required,  to  be  convenient  for  logarithmic  computation. 


40     .  PLANE    TRIGONOMETRY 

Thus,  if  sinx=i,  ^=30°,  150°,  (30° +  360°),  etc., 

and  sin-»i=30°,  150°,  (30°+36o°),  etc. 

Remark. — The  sine  or  cosine  of  an  angle  cannot  be  less  than  —  i 
or  greater  than  +  i;  hence  sin~'a  and  cos~'a  have  no  meaning  unless 
a  is  between  — i  and  +  1.  In  a  similar  manner  we  see  that  sec-'a 
and  csc~'a  have  no  meaning  if  a  is  between  —  i  and  •\- 1. 

EXERCISES 

41,  (I.)  Find  the  following  angles  in  degrees: 

sin-'^-\/2.  tan-'(— 0.  sin~*(— |)l 

COS~'^,  COS"' I, 

(2.)  If  ;r  =  cot-4>  fi"'^  tan  jr. 
(3.)  If  ;r  =  sin~'|,  find  cos.r  and  tan  x. 
(4.)  Find  sin  (tan-*^ -y^). 
(5.)  Find  sin(cos-'l). 
(6.)  Find  cot  (tan-' ^V)- 

(7.)  Given  sin~'a  =  2  cos-'a,  and  both  angles  acute;  find  a. 
(8.)  Given  sin-'a  =  cos~'a  ;  find  the  values  of  sin-'a  less  than  360"^. 
(9.)  Given  tan~'l  =|^tan-'o,  and  both  angles  less  than  360°;   find 
the  angles. 

(10.)  Given  sin~'a  =cos-'rt  and  swc^a  +  cos-'^  =450°;  find  sin-'rt. 

(II.)  Prove  sin  (cos-'a)=:=h  V^  — ^'- 

Hint.—  Let  jr=cos~'rt ;  then  a=cc)SJr, 

sin  x=-  ±  y  I  —  cos'^jr  =  ±  y  i  — a*. 

(12.)  Prove  tan(tan-'a-ftan-»^)  =  -^;— 7- 

a  —  b 
(13.)  Prove  tan  (tan-'^  —  tan~'^)=:     ,     ,» 

(14.)  Prove  cos (2  cos-'a)  =  2«'  — I. 

(1 5.)  Prove  sin (2  cos~'rt)=  d:  2a  \/i  — a*. 

2a 


I— a" 
l-a^ 


(16.)  Prove  tan  (2  tan—'a)= 

(17.)  Prove  cos(2tan~'a)= 

I  -j-a 

(18.)  Prove  sin (sin-'a4-cos-'<J)  =  «^±'v/(i— «")(!— ^ 


CHAPTER  IV 

THE    OBLIQUE    TRIANGLE 

DERIVATION  OF  FORMULAS 

42,  The  formulas  derived  in  this  and  the  succeeding 
articles  reduce  the  solution  of  the  oblique  triangle  to  its 
simplest  form. 

ft  0  C 


KIG.  2 


Draw  the  perpendicular  CD.    Let  CD=h, 

Then  ^=sin^; 


and 


(In  Fig.  2  T=sin(i8o°— .<4)=ssini4) 
h       .     J, 
a 


(In  Fig.  3  -=sin(i8oO-.ff)=8mA) 

a 

By  division  we  obtain, 


sin^ 


(32) 


h     sin  .B 

Remark. — This  formula  expresses  the  fact  that  the  ratio  of  two  sides  of  an 
oblique  triangle  is  equal  to  the  ratio  of  the  sines  of  the  angles  opposite,  and 
does  not  in  any  respect  depend  upon  which  side  has  been  taken  as  the  base. 
Hence  if  the  letters  are  advanced  one  step,  as  shown  in  the  figure,  we  obtain, 
as  another  form  of  the  same  formula, 


PLANE    TRIGONOMETRY 


.6 


b  _si.nB 

c      sinC 
Repeating  the  process,  we  obtain 

c sin  C 

a      sini4  ° 

The  same  procedure  may  be  applied  to  all  the  formulas  for  the  solution  of 

oblique  triangles.      Henceforth  only  one  expression  of  each  formula  will  be  given. 

Formula  (32)  is  used  for  the  solution  of  triangles  iji  which 
a  side  and  two  angles,  or  two  sides  and  an  angle,  opposite  one 
of  them  are  given. 

43,  We  obtain  from  formula  (32)  by  division  and  compo- 
sition, a—d__sinA-~'SinB 
a  +  l>~  sin  A  +  sin  B' 
By  formula  (29),  denoting  the  angles  by  A   and  B,  in- 
stead of  u  and  v, 

sin^-sin^     tan^(^— ^) 


Therefore, 


sin^-hsin^     tan^{A-\-B)' 
a—b     tan^(^— J5) 


(33) 


a-{-b~tan^{A  +  B)' 
This  formula  is  used  for  the  solution  of  triangles  in  which 
two  sides  and  the  included  angle  are  given. 
44,  Whether  A  is  acute  or  obtuse,  we  have 
0  C 


6  Cos  A 


(If  ^  is  acute  (Fig.  i),  AJ?  =  6  co-i  A,  DB  =  AB  -  AD  — c  -  d  cos  A,  CD- 
bsxnA.  UA  is  obtuse  (Fig.  2),  ^Z?  =  ^ cos  (180°-^)  =  -  6go%  A,  £)B=A£ 
+  AD=c-l>cosA,  CD  =  6  &in{iSo°  -  A)  =  d  sin  A .) 


THE   OBLIQUE    TRIANGLE  43 

a*={c—b  cosA)*-\-{b  sXnA)*, 
=c*—2  be  cosA  +  l^  (cosM  +sinM). 

Therefore,  a»=6*+c*-  26c  cos  A,  ^^  (34) 

This  formula  is  used  in  deriving  formula  (37). 

It  is  also  used  in  the  solution  tvithout  logarithms  of  tri- 
angles of  which  two  sides  and  the  included  angle  or  three 
sides  are  given, 

4:5,  From  formula  (34),  cos  A  = ■,"    . 

From  formula  (22),  §  37, 

b^-^-c-'-a* 


2  sm*^A  =  I  — cos^  =  I 
Hence         2sin'j^A  = 


2bc 
2bc-^a^-b'-c* 


2bc 

2bc 
_{a^b-\-c\a-\-b—c) 

"  2bc 

Let  s= ,  then  {a—b-\-c^=2{s—b),  and  (^-|-^— r)=s 

2 


2{S—C). 

Substituting,  2  sin»^^  =  ?i£z:^fcl^. 


Hence  sin^^=. /BESSES*  (35) 

V  be 

From  formula  (23),  §  37, 

2  QO^^A  =  I  +cos^  =- 


be 
2be-Vb*-V(^—(^ 


2be 
2s{s—a) 
""      be      ' 

*  In  extracting  the  root  the  plus  sign  is  chosen  because  it  is  known  that 
sin^^  is  positive. 


44 


PLANE    TRIGONOMETRY 


Hence  cos^^  =*/^fcf). 

Dividing  (35)  by  (36),  we  obtain 


(36) 

(37) 


Let 


/{s-d){s-b\s~c) 
s—av  s 


tan  ^  A 


K 

s — a 


(38) 


Formulas  (37)  and  (38)  are  used  to  find  the  angles  of  a  tri 
angle  when  the  three  sides  are  given. 

FORMULAS   FOR   THE   AREA  OF  A   TRIANGLE 
46?.  Denote  the  area  by  S. 
C 


(In  Fig.  I,  CZ>=osiniS;  in  Fig.  2,  CD-a^\xi{iZo°-B)=.a%\nB.) 

In  Figs.  I  and  2,         S=\c.CD. 

Hence  S—^ac%\xkB,  (39) 

From  formula  (17), 


THE   OBLIQUE    TRIANGLE 


45 


Substituting  for  sin^^  and  zo?>\B  the  values  found  in 
formulas  (35)  and  (36),  we  obtain 

sin  B  =  —\h{s — d){s — bts-^c), 
ac" 

Therefore,  S=k/  s{8^a'^8—h\8'—c),  (40) 

This  formula  may  also  be  written, 

S=sK.  (41) 

Formula  (39)  is  used  to  find  the  area  of  a  triangle  when 
two  sides  and  the  included  angle  are  known;  formula  (40)  or 
formula  (41),  when  the  three  sides  are  known. 

THE   AMBIGUOUS    CASE 

47.  The  given  parts  are  two  sides,  and  an  acute  angle 

opposite  one  of  them. 

Let  these  parts  be  denoted  by  a,  b,  A. 


If  a  is  less  than  b  and  greater  than  the  perpendicular  CD 
(Fig.  l),  there  are  the  two  triangles yi C'^  and  ACS',  which 
contain  the  given  parts,  or,  in  other  words,  there  are  two 
solutions. 

If  a  is  greater  than  b  (Fig.  2),  there  is  one  solution. 

If  a  is  equal  to  the  perpendicular  CD,  there  is  one  solu- 
tion, the  right  triangle  ACD. 


46  PLANE    TRIGONOMETRY 

If  the  given  value  of  a  is  less  than  CD,  evidently  there 

can  be  no  triangle  containing  the  given  parts. 

Since  CD-=b%\xiA,  there  is  no  solution  when  a<  ^sinyi  ;  there  is  one 
solution,  the  right  triangle  A  CD  when  <i=i^sin/f ;  there  are  two  solutions 
when  a  <  ^  and  >  3  sin  ^. 

4:S.  Case  I. — Given  a  side  and  two  angles, 

EXAMPLE 
Given         a  =  36.738.  A  =  36°  55'  54",  B  =  72®  5'  56", 

C=iScP—{A  +  B)z=i8oP—iogpi'  5o"  =  7oP  58' IO^ 


To^find  b. 

To  find  c. 

b      'savB 

c      sinC 

a     wa.A 

a     waA 

log  0=1.56512 

logo=l.565i2 

1<^  sin  5=9. 97845  — 10 

log  sin  C=9. 97559- 

-10 

cologsin^  =0.22123 

colog  sin  A  =0. 22123 

log  <J= 1. 76480 

log  ^^1.76194 

^5=58.184 

^=57.80 

Check. 
Determine  b  from  c,  C,  and  B  by  the  formula 
b-a_ta.n\{B-A) 
b-ira~ia.n\i^B+Ay 

This  check  is  long,  but  is  quite  certain  to  reveal  an  error.     A  check  which  is 
shorter,  but  less  sure,  is 

b  _  sin  B 

^~sinC 

Solve  the  following  triangles : 
(I.)  Given  a  =  567.25,  ^  =  11°  15',  .5  =  47°  12'. 
(2.)  Given  a  =  783.29,  A  =  81°  52',  ^  =  42°  27'. 
(3.)  Given  r=  1 125.2,  A  =79°  15',  -^  =  55°  n'. 
(4.)  Given  ^=15.346,  5=15°  51',  C=58o  10'. 
(5.)  Given  a  =  5301. 5,  A=6g°  /^',  C=4io  18'. 
(6.)  Given  ^=;  1002.1,  ^  =48°  59',  €  =  76°  3'. 

49*  Case  II. — Given  two  sides  of  a  triangle  and  the  angle 
opposite  one  of  them. 


THE   OBLIQUE    TRIANGLE 


M 


EXAMPLE 
Given  a  =  23.203.  b  =  35.121,  A  =  36°  8'  10". 

C 


^  <'' 

v/ 

'              \^ 

A 

._J ^.^^ 

To  find  B  and  B' . 

To  find  c  and  ^ , 

%\nB     b 
%\viA     a 

c     sin  (7 
<j~sin.<4 

log  '5=i.5.j556 

log  0=1.36555 

log  sin  // =9. 77064  — 10 

log  sin  C=9.9g42i- 

-10 

colog  rt=8.63445  — 10 

colog  sin ^  =  0. 22936 

log  sin  y9=9. 95065  — 10 
^=63°  12' 

log  f=  1. 58912 
^=38.825 

ff'=i800-i5=ii60  48' 

log  a  =  1.36555 

To  find  C  and  C . 

log  sin  C"  =9. 65800- 
colog  sin /4  =0.22936 

-lO 

C  =l8o°-(/f  +^)=8o°  39 

50" 

log  r' =  1.25291 

C'  =  i8o°-(^  +  Z?')=270  3- 

50" 

^'  =  17.902 

Check. 

Determine  b  from  c,  C,  and  B  by  the  formula 

b~a_isLVi\{B-A) 

d-\-a~i2Ln\{B-\-Ay 

This  check  is  long,  but  is  quite  certain  to  reveal  an  error.     A  check  whicli  is 

shorter,  but  less  sure,  is 

b_s.\nB 
c     sin  C 

(I.)  How  many  solutions  are  there  in  each  of  the  following? 

(I.)  ^==30°,  <z  =  15,  (5zr2o; 
(2.)  A  =  30°,  a  =  10,  ^  =  20; 
(3.)  B  =  30°  a  =  8,  (5  =  20 ; 
(4.)  B  =  2>7°  23'.  a  =  g.i,  b  —  7.^. 


48 


PLANE  TRIGONOMETRY 


Solve  the  following  triangles,  finding  all  possible  solutions : 
d:  =  0.63735,  ^  =  0.34312. 
«=  1.7424,  (^  =  0.96245. 
a  =  45.693,  b^  56-723- 
d!  =  9.5124,  b=  12.752. 
d!  =  0.78727,  ^  =  0.47311. 

Given  two  sides  and  the  included  angle. 


(2.)  Given  A  = 

147°  12', 

(3.)  Given  A  = 

24° 31' 

(4.)  Given  A  = 

21°  21' 

(5.)  Given  A  = 

61°  16' 

(6.)  Given  C—- 

22°32' 

SO,  Case  I)  I. 

—  Givei 

EXAMPLE 

Given  <z  =  41.003,  ^  =  48.718,  C=  68°  33' 58";  find  the  remain- 
ing parts  and  the  area. 


To  find  A  and  B. 

To  find  c. 

ta.n\{B-A)      b-a 

c     sinC 

tan^(J5+^)      b-\-a 

a     sin  A 
log  a  =  1. 61 281 

b-a=    7.715 

log  sin  C  =  9.96888— ic 

b+a  =  89.721 

colog  sin^  =0.12535 

i(5+J)=55°43'i". 

logc  =  1.70704 
c=    50.938 

log  (b-a)  =  0.88734 

To  find  the  area. 

c»Iog  (b+a)  =8.04710—10 

5  =  §  ai  sin  C 

log  tan  ^  {B+A )  =  0.16639 

logs  =  9.69897—10 

log  tan  ^(B  — .4)  =9.10083  —  10 

log  a  =  1. 61 281 

UB-A)  =    7°  11'  20" 

log  i  =  1 .68769 

hiB+A)  =  55° 43'    i" 

log  sin  C  =  9.96888— 10 

5  =  62°  54'  2l" 

log  5  =  2.96835 

4  =48°  31' 41" 

S=   929.72 

Chi 

'Xk. 

sinC     t 

z^ 

sin  5     / 

V 

log  sin  B  =  < 

J-9495I 

-10 

Iogc  = 

r. 70704 

colog  b  = 

8.31231 

—  lO 

log  sin  C  =  9.06886—10 


THE  OBLIQUE    TRIANGLE 


49 


Solve  the  following  triangles,  and  also  find  their  areas : 

(I.)  Given  A=  41°  15',  /5=o.i4726,  c=.o.\o^T\. 

(2.)  Given  C=  58°  47', /J= 11.726,    ^=16.147. 

(3,)  Given  B-=.  49°  50',  a  =  103.74,    ^=99.975. 

(4.)  Given  A=  33°  31',  ^=0.32041,  ^=0.9203. 

(5.)  Given   C=I28°   7',  (J=  17.738,    a=6o.57l. 

51,  Case  IV. — Given  the  three  sides, 

EXAMPLE 
Given  a  =  32.456,  ^  =  41.724,  ^  =  53.987  ;  find  the  angles  and  area. 


s  =  64.084 
(j  —a)  =  31.628 
{s  —  b)-=  22.360 
{s  —  c)  —  10.097 


log  (j'-rt) =1.50007 
log  (j-^)  =1.34947 
log  (j— r)=i.oo4i9 
colog  j^=8. 19325  — 10 

2)2.04698 
log  A'=  1.02349 

To  find  A. 

tan*/J= 

s—a 

log  A'=  1.02349 


log  (j— a)=i.5ooo7 


sub. 


To  find  B. 

Xaca\B= — -.' 

s—b 

log  A'=  1. 02349 
log  (j-/5)  =1.34947 


sub. 


log  tan  1^5=9.67402  — 10 
\B=2'i°  16'  16" 
^=50°  32'  32" 


To  find  C* 

tanAC= 

■*        s—c 

log  A'=  1.02349 

log  (j—f)=i. 00419 


sub. 


log  tan^C=o.oi930 

iC=46°  16'  22" 
C=9a*  32'  44" 


Ic^  tan^./?^9. 52342  — 10 
i^  =  i8°  27'  23" 
^=36°  54  46" 

Chfck. 
(^+v9  + 0=18000'  2". 

Find  the  angles  and  areas  of  the  following  triangles: 
(I.)  Given  «  =  38.5i6,  (^=44.873,  ^=14.517. 
(2.)  Given  a  =  2.ii$2>,  1>=2,-S^SA,  ^=Z-S^79- 

*  C  could  be  found  from  {A  +  j9)=(i8o°—  C),  but  for  the  sake  of  the  check  it 
is  worked  out  independently. 
4 


50  PLANE    TRIGONOMETRY 

(3.)  Given  ^=82.818,  <J=99.87i,  ^=36.363. 
(4.)  Given  ^=36.789,  <5=i  1.698,  <r=33.328. 
(5.)  Given  ^=113.08,  (^=131.17,  ^=114.29. 
(6.)  Given  a=  ,9763, /J=  1.2489,  ^=1,6543. 

EXERCISES 

52.  (I.)  A  tree,  ^,  is  observed  from  two  points,  B  and  C,  1863  ft. 
apart  on  a  straight  road.  The  angle  BCA  is  36°  43',  and  the  angle 
CBA  is  57°  21'.  Find  the  distance  of  the  tree  from  the  nearer 
point. 

(2.)  Two  houses,  A  and  B,  are  3876  yards  apart.  How  far  is  a  third 
house,  C,  from  A,  if  the  angles  ABC  and  BAC  are  49°  17'  and  58"  18' 
respectively  ? 

(3.)  A  triangular  lot  has  one  side  285.4  ft.  long.  The  angles  adja- 
cent to  this  side  are  41°  22'  and  31°  19'.  Find  the  length  of  a  fence 
around  it,  and  its  area. 

(4.)  The  two  diagonals  of  a  parallelogram  are  8  and  10,  and  the 
angle  between  them  is  53°  8' ;  find  the  sides  of  the  parallelogram. 

(5.)  Two  mountains,  A  and  B,  are  9  and  13  miles  from  a  town,  C; 
the  angle  ACB  is  71°  36'  37".  Find  the  distance  between  the  moun- 
tains. 

(6.)  Two  buoys  are  2789  ft.  apart,  and  a  boat  is  4325  ft.  from  the 
nearer  buoy.  The  angle  between  the  lines  from  the  buoys  to  the 
boat  is  16°  13'.  How  far  is  the  boat  from  the  farther  buoy?  Are 
there  two  solutions  ? 

(7.)  Given  a  =  64.256,  <:=  19.278,  C=i6°  19'  u";  find  the  differ- 
ence in  the  areas  of  the  two  triangles  which  have  these  parts. 

(8.)  A  prop  13  ft.  long  is  placed  6  ft.  from  the  base  of  an  embank- 
ment, and  reaches  8  ft.  up  its  face ;  find  the  slope  of  the  embank- 
ment. 

(9.)  The  bounding  lines  of  a  township  form  a  triangle  of  which  the 
sides  are  8.943  miles,  7.2415  miles,  and  10.817  miles;  find  the  area 
of  the  township. 

(10.)  Prove  that  the  diameter  of  a  circle  circumscribed  about  a 
triangle  is  equal  to  any  side  of  the  triangle  divided  by  the  sine  of  the 
angle  opposite. 


THE  OBLIQUE    TRIANGLE 


51 


Hint. — By  Geometry,      angle  ^4  0^=2  C. 

Draw  OD  perpendicular  to  AB, 

Angle  DOB=\AO£=C. 

DB=r  sin DOB=rwa.C. 

Hence  f=2rsinC, 

c 
or  9r=-r—p,. 


(II.)  The  distances  AB,  BC,  and  AC,  between  three  cities,  A,  B, 
and  Care  12  miles,  14  miles,  and  17  miles  respectively.  Straight  rail- 
roads run  from  A  to  B  and  C.     What  angle  do  they  make  ? 

(12.)  A  balloon  is  directly  over  a  straight  road,  and  between  two 
points  on  the  road  from  which  it  is  observed.  The  points  are  15847 
ft.  apart,  and  the  angles  of  elevation  are  found  to  be  49°  12'  and 
53°  29'  respectively.  Find  the  distance  of  the  balloon  from  each  of 
the  points. 

(13.)  To  find  the  distance  from  a  point  ^  to  a  point  B  on  the  op- 
posite side  of  a  river,  a  line,  AC,  and  the  angles  CAB  and  ACB  were 
measured  and  found  to  be  315.32  ft.,  58°  43',  and  57°  13'  respectively. 
Find  the  distance  AB. 

(14.)  A  building  50  ft.  high  is  situated  on  the  slope  of  a  hill.  From 
a  point  200  ft.  away  the  building  subtends  an  angle  of  12°  13'.  Find 
the  distance  from  this  point  to  the  top  of  the  building. 

(15.)  Prove  that  the  area  of  a  quadrilateral  is  equal  to  one-half 
the  product  of  the  diagonals  by  the  sine  of  the  angle  between 
them. 

(16.)  From  points  A  and  B,  at  the  bow  and  stern  of  a  ship  respec- 
tively, the  foremast,  C,  of  another  ship  is  observed.  The  points  A 
and  B  are  300  ft.  apart ;  the  angles  ABC  and  BAC  are  found  to  be 


52  PLANE    TRIGONOMETRY 

65°  31' and  110°  46' respectively.     What  is  the  distance  between  the 
points  A  and  C  of  the  two  ships  ? 

(17.)  Two  steamers  leave  the  same  port  at  the  same  time  ;  one  sails, 
directly  northwest,  12  miles  an  hour;  the  other  17  miles  an  hour,  in 
a  direction  67°  south  of  west.  How  far  apart  will  they  be  at  the  end 
of  three  hours  ? 

(18.)  Two  stakes, .(4  and  B,a.rt  on  opposite  sides  of  a  stream;  a 
third  stake,  C,  is  set  62  ft.  from  A ;  the  angles  ACB  and  CAB  are 
found  to  be  50°  3'  5"  and  61°  18'  20"  respectively.  How  long  is  a 
rope  connecting  A  and  B} 

{19.)  To  find  the  distance  between  two  inaccessible  mountain-tops, 
A  and  B,  of  practically  the  same  height,  two  points,  C  and  D,  are 
taken  one  mile  apart.  The  an,!j;le  CDA  is  found  to  be  88°  34',  the 
angle  DCA  is  63''  8',  the  angle  CDB  is  64°  27',  the  angle  DCB  is  87"  9'. 
What  is  the  distance? 

(20.)  Two  islands,  B  and  C,  are  distant  5  and  3  miles  respectively 
from  a  light-house,  .(4,  and  the  angle  BAC  is  33°  11';  find  the  dis^ 
tance  between  the  islands. 

(21.)  Two  points,  A  and  B,  are  visible  from  a  third  point  C,  but 
not  from  each  other;  the  distances  AC,  BC,  and  the  angle  ACB  were 
measured,  and  found  to  be  1321  ft.,  1287  ft.,  and  61°  22'  respectively. 
Find  the  distance  AB. 

(22.)  Of  three  mountains.  A,  B,  and  C,  B  is  directly  north  of  C  5 
miles,  ^  is  8  miles  from  C  and  1 1  from  B.     How  far  is  A  south  oi  B? 

(23.)  From  a  position  215.75  ft.  from  one  end  of  a  building  and 
198.25  ft.  from  the  other  end,  the  buildmg  subtends  an  angle  of 
53°  37'  28";  find  its  length. 

(24.)  If  the  sides  of  a  triangle  are  372.15,  427.82,  and  404.17  ;  find 
the  cosine  of  the  smallest  angle. 

(25.)  From  a  point  3  miles  from  one  end  of  an  island  and  7  miles 
from  the  other  end,  the  island  subtends  an  angle  of  33°  55'  15";  find 
the  length  of  the  island. 

(26.)  A  point  is  13581  in.  from  one  end  of  a  wall  12342  in.  long,  and 
10025  '"•  from  the  other  end.  What  angle  does  the  wall  subtend  at 
this  point? 

(27.)  A  straight  road  ascends  a  hill  a  distance  of  213.2  ft.,  and  is  in- 


THE  OBLIQUE   TRIANGLE  53 

clined  12"  2'  to  the  horizontal ;  a  tree  at  the  bottom  oX  the  hill 
subtends  at  the  top  an  angle  of  10°  5'  16".  Find  the  height  of  the 
tree. 

(28.)  Two  straight  roads  cross  at  an  angle  of  37°  50'  at  the  point  A ; 
3  miles  distant  on  one  road  is  the  town  B,  and  5  miles  distant  on  the 
other  is  the  town  C.     How  far  are  B  and  C  apart? 

(29.)  Two  stations,  A  and  B,  on  opposite  sides  of  a  mountain,  are 
both  visible  from  a  third  station,  C;  ^C=  11.5  miles,^C=94  miles, 
and  the  angle  ACB  =  $(^  31'.     Find  the  distance  from  A  to  B. 

(30.)  To  obtain  the  distance  of  a  battery,  A,  from  a  point,  B,  of  the 
enemy's  lines,  a  point,  C,  2i7^-7  yards  distant  from  A  is  taken ;  the  an- 
gles ACB  and  CAB  are  measured  and  found  to  be  7gP  53'  and  74°  35* 
respectively.    What  is  the  distance  AB} 

(31.)  A  town,  B,  is  14  miles  due  west  of  another  town,  A.  A  third 
town,  C,  is  19  miles  from  A  and  17  miles  from  B.  How  far  is  Cwest 
of^? 

(32.)  Two  towns,  A  and  B,  are  on  opposite  sides  of  a  lake.  A  is 
18  miles  from  a  third  town,  C,  and  B  is  13  miles  from  C;  the  angle 
ACB  is  13°  17'.     Find  the  distance  between  the  towns  A  and  B. 

(33.)  At  a  point  in  a  level  plane  the  angle  of  elevation  of  the  top 
of  a  hill  is  39°  51',  and  at  a  point  in  the  same  direct  line  from  the  hill, 
but  217.2  feet  farther  away,  the  angle  of  elevation  is  26°  53'.  Find 
the  height  of  the  hill  above  the  plane. 

(34.)  It  is  required  to  find  the  distance  between  two  inaccessi- 
ble points,  A  and  B.  Two  stations,  C  and  D,  2547  ft.  apart,  are 
chosen  and  the  angles  are  measured ;  they  are  ACB=^2'j°  21',  BCD 
=33°  14',  BDA=id>°  17',  and  ADC=S^°  23'.  Find  the  distance  from 
^  to^. 

(35.)  Two  trains  leave  the  same  station  at  the  same  time  on  straight 
tracks  inclined  to  each  other  21°  12',  If  their  average  speeds  are  40 
and  50  miles  an  hour,  how  far  apart  will  they  be  at  the  end  of  the  first 
fifteen  minutes  ? 

(36.)  A  ship,  A,  is  seen  from  a  light-house,  B;  to  determine  its  dis- 
tance a  point,  C,  300  ft.  from  the  light-house  is  taken  and  the  angles 
BCA  and  CBA  measured.  If  BCA  =  108°  34'  and  CBA  =65®  27',  what 
is  the  distance  of  the  ship  from  the  light-house  ? 


54 


PLANE  TRIGONOMETRY 


(37.)  Prove  that  the  radius  of  the  inscribed  circle  of  a  triangle  is 
equal  to  a  sin^^  sin  |^Csec  \A. 


Hint. — Draw  OR,  OC,  and  the  perpendicular  OD. 
OB  and  OC  bisect  the  angles  B  and  C  respectively,  and  OD=r. 
a  =  BD-\-DC=r{coi\B  +  co\.\  C). 
coi\B  +  cot  i C=  ^'"Kcosj^  +  cosiCsin^jg^ 
sin  ^  B  sin  \  C 
_  sin^(Z?+C)  _        cosi^ 
sin  I  B  sin 
sin  4  ^  sin  A  C 


Hence 


_  C     sin  5  ^  sin  ^  C 
—  a  sin  .]  ^  sin  J  C  sec  \  A. 


CHAPTER  V 


CIRCULAR  MEASURE— GRAPHICAL  REPRESENTATION 


CIRCULAR   MEASURE 


53,  The  length  of  the  semicircumference  of  a  circle  is 
7r^(7r  =  3.i4i59+);  the  angle  the  semicircumference  sub- 
tends at  the  centre  of  the  circle  is  i8o°.  Hence  an  arc 
whose  length  is  equal  to  the  radius  will  subtend  the  angle 

jOqO 

',   this  angle  is   the   unit   angle  of    circular   measure, 


and  is  called  a  radian. 


nR 


If  the  radius  of  the  circle  is  unity,  an  arc  of  unit  length 
subtends  a  radian ;  hence  in  the  unit  circle  the  length  of  an 
arc  represents  the  circular  measure  of  the  angle  it  subtends. 

Thus,  if  the  length  of  an  arc  is  -,  it  subtends  the  angle  -  radians. 

Since  one  radian  = ,  we  have 

TT 

90°  =  —  radians, 
180°=  TT  radians, 


56 


PLANE    TRIGONOMETRY 
270°=  —  radians, 


360°  =  27r  radians,  etc. 
The  value  of  a  radian  in  degrees  and  of  a  degree  in  radians  are ; 
I  radian  =  57.29578°, 

=  57°  17' 45"- 
i°=. 0174533  radian. 
In  the  use  of  the  circular  measure  it  is  customary  to  omit  the  word  radian  \ 

thus  we  write  -  ,  v,  etc.,  denoting  -  radians,  ir  radians,  etc.      On  the  other 
2  *•  2 

hand,  the  symbols  o  '  "  are  always  printed  if  an  angle  is  measured  in  degrees, 

minutes,  and  seconds ;  hence  there  is  no  confusion  between  the  systems. 


EXERCISES 

(I.)  Express  in  circular  measure  30°,  45°  60°,  120°  135°,  720°,  990®. 
(Take  7r  =  3.i4i6.) 

(2.)  Express  in  degrees,  minutes,  and  seconds  the  angles  ^,  —  ,- ,-. 

(3.)  What  is  the  circular  measure  of  the  angle  subtended  by  an  arc 
of  length  2.7  in.,  if  the  radius  of  the  circle  is  2  in..?  if  the  radius  is 
5  in.? 

54:*  The  following  important  relations  exist  between  the 
circular  measure  x  of  an  angle  and  the  sine  and  tangent  of 
the  angle. 

(i.)  If  X  is  less  than  — ,  sinx<x <tanx. 


Draw  a  circle  of  unit  radius. 

By  Geometry,        SP <z.xc  AP <AT. 
Hence  sin;ir<jir<  tan;tr. 


CIRCULAR  MEASURE  57 

siti  X         tttfi  X 
(2.)  As  X  approaches  the  limit  o,  — —  and  approach 

the  limit  i. 

Dividing  sin  ;r  <  ;r  <  tan  x  by  sin  x,  we  obtain 

X  I 

\<-. < 


sin;r     Qosx 

sinjf     cos;p 
I  n verting,  i  >  —^  >  "y-  . 

As  X  approaches  the  limit  o,  cos;r  approaches  the  length 

of  the  radius,  that  is,  i,  as  a  limit. 

sin  X 
Therefore, approaches  the  limit  i. 

Dividing  i  > >  cosjr  by  cos;r,  we  obtain 

I         tan;r 

>— 7->I. 


COs;r         X 
As  X  approaches  the  limit  o,  cos;ir  approaches  the  limit  i ; 

hence approaches  the  limit  I. 

cos  X    ^^ 

tan  X 
Therefore,         • approaches  the  limit  i. 


PERIODICITY  OF  THE  TRIGONOMETRIC  FUNCTIONS 

SS*  The  sine  of  an  angle  x  is  the  same  as  the  sine  of 
(;r 4-360°),  {x-^720°),  etc. — that  is,  of  {x+2n'jr),  where  «  is 
any  integer. 

The  sine  is  therefore  said  to  be  a  periodic*  function,  hav- 
ing the  period  360°,  or  27r. 

The  same  is  true  of  the  cosine,  secant,  and  cosecant. 

*  If  a  function,  denoted  by/(jr),  of  a  variable  x,  is  such  that  /{x+k)^/{x) 
for  every  value  of  x,  k  being  a  constant,  the  function  f{x)  is  periodic ;  if  ii  is 
the  least  constant  which  possesses  this  property,  k  is  the  period  of  f{x). 


58 


PLANE    TRIGONOMETR  V 


The  tangent  of  an  angle  x  is  the  same  as  the  tangent  of 
{x-\- 1 80°),  (;r4-36o°),  etc. — that  is,  of  {x-\-nir\  where  n  is  any 
integer. 

The  tangent  is  therefore  a  periodic  function,  having  the 
period  1 80°,  or  tt. 

The  same  is  true  of  the  cotangent. 


GRAPHICAL    REPRESENTATION 

5Q,  On  the  Hne  OX  lay  off  tlie  distance  OAi^—x)  to  rep- 
resent the  circular  measure  of  the  angle  x.  At  the  point  A 
erect  a  perpendicular  equal  to  sin  x.  If  perpendiculars  are 
thus  erected  for  each  value  of  x,  the  curve  passing  through 
their  extremities  is  called  the  sine  curve. 

If  sin X  is  negative,  the  perpendicular  is  drawn  downward. 


In  a  similar  manner  the  cosine,  tangent,  cotangent,  secant, 
and  cosecant  curves  can  be  constructed. 


+1 


-1 


+1 


/ 

1 

1 

1 

\ 

/ 

I 

1 

\ 

0 

% 

\ 

/2  7r 

^TT 

3 

Sine  Curve 


^ 

\ 

1 
1 

1 

\ 

X 

0 

7^ 

1 
V^    1   ^ 

y/Vi^ 

27r 

g^'rX 

3^ 
1 

Cosine  Curve 


GRA  PHICAL    REP  RES  EN  TA  TION 


59 


Cotangent  Curve 


6o 


PLANE    TRIGONOMETRV 


SBCANT  CURVE 


If  the  distances  on  OX  are  measured  from  Cy  instead  of 
O,  we  obtain  from  the  secant  curve  the  cosecant  curve. 

In  the  construction  of  the  inverse  curves  the  number  is 
represented  by  the  distance  to  the  right  or  left  from  O; 
the  circular  measure  of  the  angle  by  the  length  of  the  per- 
pendicular erected. 

All  of  the  preceding  curves,  except  the  tangent  and  co- 
tangent curves,  have  a  period  of  27r  along  the  line  OX;  that 
is,  the  curve  extended  in  either  direction  is  of  the  same 
form  in  each  case  between  27r  and  47r,  477  and  &ir, — 27r  and 
o,  etc.,  as  between  o  and  277,  while  the  corresponding  inverse 
curves  repeat  along  the  vertical  line  in  the  same  period. 
The  period  of  the  tangent  and  cotangent  curves  is  tt. 


GRAPHICAL   REPRESENTATION  61 


-1  0         +1 


-1  0  +, 


INVBRSB    SINE   CURVB 


INVERSE   COSINE   CURVB 


-2  -1  0  +1  +2  +3 


INVERSE   TANGENT   CURVE 


62 


PLANE    TRIGONOMETRY 


a^JT 


-3 


-2 


-I 


+  i 


+  2         +3 


I.NVSKbS    SECANT 


CHAPTER   VI 

COMPUTATION  OF  LOGARITHMS  AND  OF  THE  TRIG- 
ONOMETRIC FUNCTIONS -DE   MOIVRE'S  THEOREM 
—HYPERBOLIC   FUNCTIONS 

St*  A  convenient  method  of  calculating  logarithms  and 
the  trigonometric  functions  is  to  use  infinite  series.  In 
works  on  the  Differential  Calculus  it  is  shown  that 

jp2        /j»8        /]«j4 

iogr,(l+a5)  =  a5-^+^-^+...  (I) 

/y»3         rpf>         rgtt 

iliia5=a5-^+^j-yj+...*  (a) 

,     a;2     a;*     a5« 
COia5=l-2-j  +  4j-^j  +  ...  (3) 

Another  development  which  we  shall  use  later  is 

where  ^=2.7182818  ...  is  the  base  of  the  Naperian  system 
of  logarithms. 

58,  The  series  (i)  converges  only  for  values  of  x  which  satisfy  the 
inequality  —  i<x^i.  The  series  (2),  (3),  and  (4)  converge  for  all 
finite  values  of  x. 

It  is  to  be  noted  that  the  logarithm  in  (i)  is  the  Naperian,  and  the 
angle  x  in  (2)  and  (3)  is  expressed  in  circular  measure. 

*  3!  denotes  1x2x3;  41  denotes  1x2x3x4,  etc. 


64  PLANE   TRIGONOMETRY 

COMPUTATION    OF   LOGARITHMS 

59*  We  first  recall  from  Algebra  the  definition  and  some 
of  the  principal  theorems  of  logarithms. 

The  logarithm  to  the  base  a  of  the  number  m  is  the  number  x 
which  satisfies  the  equation. 

This  is  written  x  =  log^  m. 

The  logarithm  of  the  product  of  two  numbers  is  equal  to  the  sum 
of  the  logarithms  of  the  numbers. 

Thus  log^  mn  =  log^  w  +  log^  n. 

The  logarithm  of  the  quotient  of  two  numbers  is  equal  to  the  log- 
arithm of  the  dividend  minus  the  logarithm  of  the  divisor. 

Thus  loga-  =  logtfW  — log^«. 

n 

The  logarithm  of  the  power  of  a  number  is  equal  to  the  logarithm 
of  the  number  multiplied  by  the  exponent. 

Thus  log^  m^=p  log^  m. 

To  obtain  the  logarithm  of  a  number  to  any  base  a  from  its  Na- 

perian  logarithm,  we  have 

log^  m 

log*  f^  = =  M^  log,  m, 

log*« 

where  M-  = . ;  M .  is  called  the  modulus  of  the  system. 

log,  a 

60»  We  proceed  now  to  the  computation  of  logarithms. 

The  series  (i)  enables  us  to  compute  directly  the  Naperian 

logarithms  of  positive  numbers  not  greater  than  2. 

•J 
Example. — To  compute  log^-  to  five  places  of  decimals. 

2 
Substitute  -  for  jr  in  (i): 

2 

3,        /,i\iii,ii       II, 

If  the  result  is  to  be  correct  to  five  places  of  decimals,  we  must  take  enough 
terms  so  that  the  remainder  shall  not  affect  the  fifth  decimal  place.     Now  we 


COMPUTATION  OF  LOGARITHMS 


65 


know  by  Algebra  that  in  a  series  of  which  the  terms  are  each  less  in  nnmerical 
value  than  the  preceding,  and  are  also  alternately  positive  and  negative,  the  re- 
mainder is  less  in  numerical  value  than  its  first  term.  Hence  we  need  to  take 
enough  terms  to  know  that  the  first  term  neglected  would  not  affect  the  fifth 
place. 


Positive  terms 

I 
2  ~ 

>.  5000000 

I 
3 

I 
■55- 

.0416667 

I 
5 

I 

.0062500 

I 

7 

I 

.OOIII6I 

I 
9 

I 

'2«~ 

.0002170 

I 
II 

I 
"2"" 

.0000444 

I 
13 

X 

.0000094 

.5493036 


Negative  terms 

I 

2 

I 

-j  =0.1250000 

I 
4 

I 

.0156250 

I 
6 

I 
2«~ 

.0026042 

I 
8 

I 

28  ~ 

.0004883 

I 
10 

I 

.0000977 

I 
12 

I 

?2~ 

.0000203 

I 
14 

I 

.0000044 

•1438399 


Subtracting  the  sum  of  the  negative  from  the  sum  of  the  positive  terms,  we 
obtain 

log^|  =  •4054637- 
Denote  the  sum  of  the  remaining  terms  of  the  series  by  R.     Then,  by  Alge- 
bra, 

15     2>* 
< .0000021. 

The  error  caused  by  retaining  no  more  decimal  places  in  the  computation  is 
less  than  .0000006.  Hence  the  total  error  is  less  than  .0000027.  Therefore 
the  result  is  correct  to  five  decimal  places. 

SI,  As  remarked,  the  series  (i)  does  not  enable  us  to 
calculate  directly  the  logarithms  of  numbers  greater  than  2, 
but  it  can  be  readily  transformed  into  a  series- which  gives 
us  the  logarithm  of  any  positive  number. 

Replacing  x\>y  —  ^  in  (i),  we  obtain 
5 


66  PLANE    TRIGONOMETRY 

X*        X*       X* 

log,  (i  —  x)=  —  x —  .  ,  . 

234 

This  series  converges  for  —  i^j;<i. 

Subtracting  this  from  (i),  we  obtain 

log,  (i-f  ;r)-log,  (i-x)  =  log,  (j^) 

which  converges  for  —  i  <  ^  <  i . 

Putting  _y=  I 1,  we  see  that  j  passes  from  o  to  00  as  :»; 

passes  from  —  i  to  +1  ;  hence,  if  we  make  this  substitution  in 
(5),  we  get  a  series 

which  converges  for  all  positive  values  oi  y,  and  therefore  enables 
us  to  compute  the  Naperian  logarithm  of  any  number. 

From   (5)  we   can   get   another  series  which  is   useful :    put 

I        ,  1+^   y-{- 1  .... 

;  then,  as = ,  equation  (5)  gives  us 


27 -f  I  i—x       y 

which  converges  for  all  positive  values  oi  y.     Hence, 

log,(,+  .)  =  log,,+  .(j^  +  i.^.j^.+l.jj^+...).<7) 

This  series  gives  us  log,(_>'-f-i),  when  log^j  is  known.  It  con- 
verges more  rapidly  than  (6),  when  y  is  greater  than  2,  and  hence 
should  be  used  under  these  circumstances. 

62,  To  construct  a  table  we  need  to  compute  directly 
only  the  logarithms  of  prime  numbers,  since  the  others  can 
be  obtained  by  the  relation 

log  ,ry  =  log  x-\-\og y. 


COMPUTATION  OF  LOGARITHMS 


67 


Thus,  to  obtain  the  logarithms  of  the  integers  up  to  10, 

we  need  to  compute  by  series  only  the  logarithms  of  the 

numbers  2,  3,  5,  and  7. 

(For  4=2'',  6=2.  3,  8=2*,  9=3',  ro=2  .  5,  and  log  1=0.) 
In  this  case  we  are  computing  the  logarithms  of  successive  integers,  and 
should  therefore  use  (7). 

G3,  Example. — Compute  the  Naperian  logarithms  of  2,  3,  4,  and  5. 
/I     11,11,1     I.I      I  \ 

log.  2  =  2( -  +  -.-.+      .  -  +  -    .  -  +  -.-1+   .   .   .\ 

\3    3    3'     5    3*    7    3'     9     3  / 

Denote  the  sum  of  the  remaining 
terms  of  this  series  by  R. 
Then,  by  Algebra, 

^<IT-3^-I^' 

or       i?<  .ooo(xx)573. 

The  error  caused  by  not  retaining 
more  places  of  decimals  in  the  pre- 
ceding column  is  less  than  .0000005. 
Hence,  the  total  error  is  less  than 
.00000165. 
log^2=. 6931458 

Remark. — We  should  get  the  same  series  if  we  were  to  use  (6). 


3 

.3333333 

I 

3 

I 

•0123457 

I 

I 
3^~ 

.0008230 

I 
7 

I 

.0000653 

I 

9 

I 

.0000056 

.3465729 
2 

10& 


3  =  log.2  +  2(l  +  ^-i  +  i.l+i.i+...V 

^       ^  V5     3    5      5    5*     7    5'  / 


-=.2000000 
5 

—  =  .0026667 

5' 

-^=.0000640 

—  =  .0000018 
5' 


.2027325 
2 

.4054650 
•6931458 


Add  log*  2  = 

logt  3=1.0986108 


R<'^ 
9 


or  >*?<  .00000006. 

Noting  the  errors  in  the  pre- 
ceding column  and  in  log*  2,  we 
see  that  the  total  error  is  less  than 
.000002x7. 


68  PLANE    TRIGONOMETRY 

Remark. — If  we  were  to  use  (6)  to  compute  log^3,  we  should  have 

.o.3=.[i+i©'+K0'+K^y-]- 

This  series  converges  much  more  slowly  than  the  above,  since  its 
terms  are  multiples  of  powers  of  \.  while  the  terms  of  the  above  are 
the  same  multiples  of  powers  of  \.  Thus,  we  should  be  obliged  to 
use  eight  instead  of  four  terms  to  have  the  result  correct  to  five 
places. 

log*  4=2  log*  2  =  1. 3862916. 

Iog.5  =  log.4+2Q  +  i.^+i.^.+  ...), 
or  log*  5  =  1.60944. 

G4:,  Proceeding  in  like  manner,  we  may  calculate  any  number  of 
logarithms. 

The  following  table  gives  the  Naperian  logarithms  of  the  first  ten 
integers :  • 


log^  I  =  .00000 
log,  2=  .69315 
log,  3  =  1.09861 
log«  4  =  1-38629 
log,  5  =  1-60944 


log,  6  =  1.79176 

log,  7  =  1-94591 

log,  8  =  2.07944 

log,  9  =  2. 19722 

log,  10  =  2.30259 


The  common  logarithm  of  any  number  may  be  found  by  multiply- 
ing its  Naperian  logarithm  by  M,o=:  .43429448.  §  59 

Thus  log.o  5  =  log,  5  X  .43429448  =  .69897. 

60.  Remark.— \i  a  table  of  logarithms  were  to  be  computed,  the 
theory  of  interpolation  and  other  special  devices  would  be  employed. 

COMPUTATION   OF   TRIGONOMETRIC   FUNCTIONS 

sin  f  cos  T 

as.  Since  tan,r=-^ — -,  cot  ;r—    /  '-,  etc.,  the  computa- 
cos,i'  sin;tr 

tion  of  all  the  trigonometric  functions  depends  upon  that  of 
the  sine  and  cosine;  thus  the  developments  (2)  and  (3)  suf- 
fice for  all  the  trigonometric  functions.     Further,  since  the 


COMPUTATION  OF   SINES  AND   COSINES  69 

sine  or  cosine  of  any  angle  is  a  sine  or  cosine  of  an  angle 

;:;— ,  it  is  never  necessary  to  take  x  greater  than  -  in  the 
<^4  4 

series  (2)  and  (3),  §  16 

Since  -  =0.785398  .  .  .  <^_,  these  series  converge  rapidly;  in  fact, 
4  10 

—  =  .000003  does  not  affect  the  fifth  decimal   place,  and  —  the 
9!  11! 

seventh. 

67.  Remark. — In  the  systematic  computation  of  tables  we  should 
not  calculate  the  functions  of  each  angle  from  the  series  independent- 
ly. We  should  rather  make  use  of  the  formulas  (25)  and  (27)  of  §  38, 
thus  obtaining 

smnx  =  2  cosx  sin  {n  —  i)x  —  sin  {n  —  2)x, 
cosnx-=:2  cos.r  cos(« —  i)x  —  cos(«  —  2)x. 

If  our  tables  are  to  be  at  intervals  of  i',  we  should  calculate  the 
sine  arid  cosine  of  i'  by  the  series.  The  above  expressions  then  en- 
able us  to  find  successively  the  sine  and  cosine  of  2',  3',  4',  etc.,  till  we 
have  the  sine  and  cosine  of  all  angles  up  to  30°  at  intervals  of  i'. 

To  obtain  the  sine  and  cosine  of  angles  from  30°  to  45°  we  should 
make  use  of  these  results  by  means  of  the  formulas 
sin  (30°+ J/)  :=cos,y  —  sin  (30° — y), 
cos  (30°+^)  =  cos  (30°— J/)  — s\ny. 

68*  To  employ  series  (2)  and  (3)  in  computing  the  sine 
and  cosine  we  must  first  convert  the  angle  into  circular 
measure. 

To  do  this  we  recall  that 

1°  =  .017453293,     I '  =  .0002908882,     I  "  =  .000004848 1 37. 

Example. — To  compute  the  sine  and  cosine  of  12^  15'  39". 

12°=  .209439516 
15'  =.004363323 
39"  =  .000189076 
12°  15'  39"  =  .213991915  in  circular  measure. 


70 


PLANE   TRIGONOMETRY 


sinjr  =  x -H — -— 

3!     5! 

jr=.2I399I9 
— I  =  .0000037 

.3139956 
subtract  — ^=.0016332 

sin  jr=. 2123624 
Correct  to  five  decimal  places. 


cosjr=i ■-! 

2 !      4! 

1  =  1.0000000 

— •=  .0000874 
4! 

1.0000874 
subtract  — j=  .0228963 

cosjr=  .9771911 
Correct  to  five  decimal  places. 


DE  MOIVRE'S  theorem 
69.  In  Algebra  we  learn  that  the  complex  number 

a=a+/3\/^=o+/3/  (8) 

may  be  represented  graphically  thus : 


Take  two  lines,  OX  and  OV,  at  right  angles  to  each  other. 
To  the  number  a  will  correspond  the  point  A,  whose  dis- 
tances from  the  two  lines  of  reference  are  yS  and  a  re- 
spectively. 

This  geometrical  representation  shows  at  once  that  we 
can  also  write  a  in  the  form 

a=r  (cos5  +  i  sin  3).  (9) 

70,  From  Algebra  we  recall  the  definition  of  the  sum  of  the 
complex  numbers  a  =  a  +  tj3  and  d=y  +  iS\  namely 

Subtraction  is  defined  as  the  inverse  of  addition,  so  that 
a— d=a—y -f/(/3— 3). 


DE  MOIVRE-S    THEOREM  71 

Multiplication  is  most  conveniently  defined  when  a  and  b  are 
written  in  form  (9).     If 

a~r  (cos  ^  +  / sin  ^)  and  b=s  (cos^+Zsin^), 
their  product  is  defined  by  the  equation 

«^^rj  [cos(&+^)-|-/sin(^4-0)].  (10) 

Division  is  defined  as  the  inverse  of  multiplication,  so  that 

-=-  [cos(^~<p)  +  i  sin  {^—(f)]. 

Finally,  we  recall  that  in  an  equation  between  complex  numbers, 

we  have  a=y»    /3=5.  (11) 

71*  Consider  the  different  powers  of  the  complex  number 

ic=cos.&4-«  sin^. 
By  (10)  we  have 

;«:''= (cos  ^+/  sin  S)  (cos  ^+/  sin  ^), 
=cos2^-j-/sin  2^. 
x'^x"* .  :t=(cos  2^+/  sin  2^)  (cos  ^-\-i  sin ^), 
=cos35-f-^  sin3^. 
And,  in  general,  for  any  integer  n, 

AT" = (cos  B-\-t  sin  ^)'*=cos  «■&+/  sin  n^. 
From  this  equation  we  have  De  Moivre's  Theorem,  which 
is  expressed  by  the  formula 

(co8^+isin^)"=(coi«^+i8lnn^).  (12) 

72,  An  interesting  application  of  De  Moivre's  Theorem 
is  the  expansion  of  sin  «;ir  and  cos«;r  in  terms  of  sin;ir  and 
cos;r.  Expanding  the  left-hand  side  of  (12)  by  the  bino- 
mial theorem,  and  substituting  x  for  S-,  we  have 

cos «;<:+/  sin  nx=cos''x-^n  cos""'  x  {i  sin  x")  -\-— — — ^  cos"""* ;« 

..   .  n.{n  —  i)(n—2)       ^_,     ...      ,,   , 

(t  sin  xY  H ^^ — ^  cos"    ^xCtsmx)*  -{-  .  . . 

3  ! 


72  PLANE    TRIGONOMETRY 

or 

cos«ji:-|-/  s\Xinx=\c.o^^ X ^^ — j — -  cos""':*:  sin' j;4-  , .  .1 

[«_,       •          n{n  —  i)ln  —  2)        „_,       ...         1 
n  cos"    ^xsmx ^ — . — -^ ^  cos"    ^  x  sm  x-i^  . ,  A. 
3  1  J 

Equating  real  and  imaginary  parts,  as  in  (li),  we  have 

cos«^=cos''a: ^ — j — -cos"    *xs\nx-{-...  (13) 

«     .       .  «(«  — 1)(«  — 2)  .   -  ,     . 

sin«j;zi:«cos*'    'xsin;*: ^^ — cos''~3;t;sinVx-f-. ..  (14) 

3  1 
Example. — n  =  5. 

cos  5;r  =  cos' X  —  10  cos^jr  sin';r+5  cosx  sin*;r. 

sin  5;r  =  5  cos*  ^  sin  ;r  — 10  cos"  x  sin'  x  +  sin*  x. 

THE   ROOTS   OF  UNITY 

73,  We  find  another  appHcation  of  De  Moivre's  Theorem 

jn  obtaining  the  roots  of  unity.     The  n^^  roots  of  unity  are 

by  definition  the  roots  of  the  equation 

;r*=i. 

Every  equation  has  n  roots  and  no  more ;  hence,  if  we 

can  find  n  distinct  numbers  which  satisfy  this  equation  we 

shall  have  all  the  n*^  roots  of  unity. 

Consider  the  n  numbers 

27rr      .    .     27rr 

j:^=cos \-t  sm , 

«  n 

r=.o,  1,  2,  .  .  .  n—i. 

Geometrically  these  numbers  are  represented  by  the  n 
vertices  of  a  regular  polygon.  They  are,  therefore,  all  dif- 
ferent. We  shall  see  now  that  they  are  precisely  the  n^ 
roots  of  unity. 

In  fact,  we  have  by  (12), 


,-{ 


2wr      .   .     2Tr\ 

cos \-t  sin ,  , 

n  n  j 


:V. 


THE  ROOTS   OF  UNITY  73 

=cos  25rr+/sin  2?rr, 

=:i+i.  0=1. 
Therefore  x^  is  one  of  the  roots  of  unity. 

Thus  the  cube  roots  of  unity  are  represented  by  the  points  A,  P, 
and  Q  of  the  following  figure.     In  the  figure  0A  =  i,  angle  AOP  = 

—  =  120°  angle  AOQ  =  — =  2^0°;  that  is,  the  circumference  is  di- 
vided into  three  equal  parts  by  the  points  A,  P,  and  Q.  Then  OD  =  i, 
and  DP  =  DQ  =  ^-\/-i.  Hence  we  see  from  the  method  of  represent- 
ing a  complex  number  given  above  that  ^  represents -|-i,/' represents 
— i+' i\^3.  Q  represents  —^  —  ^k^/'i• 


EXERCISES 

74,  (I.)  Express  sin  ^  and  cos/jj*:  in  terms  of  sin  jr  and  cosr. 

(2.)  Express  sin  6x  and  cos  6x  in  terms  of  sin  ^  and  cos^r. 

(3.)  Find  the  six  6*  roots  of  unity. 

(4.)  Find  the  five  5*  roots  of  unity. 


THE  HYPERBOLIC  FUNCTIONS 

75*  The  hyperbolic  functions  are  defined  by  the  equations 


sinb  X  = 


cosh  £C  = 


2      ' 


(15) 


(16) 


in  which  sinh;ir  and  cosh;r  denote  the  hyperboh'c  sine  and 


74  PLANE  TRIGONOMETRY 

hyperbolic  cosine  of  x  respectively.  These  functions  are 
called  the  hyperbolic  sine  and  cosine  on  account  of  their 
relation  to  the  hyperbola  analogous  to  the  relation  of  the 
sine  and  cosine  to  the  circle.  A  natural,  and  convenient 
way  to  arrive  at  the  hyperbolic  functions  and  to  study  their 
properties  is  by  using  complex  numbers  in  the  following 
manner.  The  series  (2),  (3),  and  (4)  give  the  value  of  sin  x, 
cos  X,  and  e'^  for  every  real  value  of  x.  These  series  also 
serve  to  define  sin  x,  cos  x,  and  t-^  for  complex  values  of  x. 
In  the  more  advanced  parts  of  Algebra  it  is  shown  that 
the  following  fundamental  formulas  which  we  have  proved 
only  for  a  real  variable, 

sin  {^x  -\- y)  =  sin  x  co?> y -\-  cos  a:  sin_y,  (17) 

cos  (a: -f- J')  =  cos  X  cos  7  — sin  x  sin_y,  (18) 

e^+y  =  e''ey,  (19) 

hold  unchanged  when  the  variable  is  complex. 

This  fact  enables  us  to  calculate  with  ease  sin  x^  cosXy  and 
e^  for  any  complex  value  of  the  variable. 

In  so  doing  we  are  led  directly  to  the  hyperbolic  functions. 
At  the  same  time  a  relation  between  the  trigonometric  and 
hyperbolic  functions  is  estabhshed  by  means  of  which  the 
formulas  of  Chapter  III.  can  be  converted  into  correspond- 
ing formulas  for  the  hyperbolic  functions. 

Taking  x  and  y  real  and  replacing  y  in  (17),  (18),  and  (19)  by 
iy,  we  get 

sin  (^x  +  iy^  =  sin  x  cos  t'y  +  cos  x  sin  ty, 
cos  (x  +  t'y)  =  cos  X  cos  t'y  —  sin  x  sin  iy, 

Thus  the  calculation  of  these  functions  when  the  variable 
is  complex  is  made  to  depend  upon  the  case  where  the  vari- 
able is  a  pure  imaginary. 


HYPERBOLIC  FUNCTIONS  7S 

If  we  replace  x  by  ix  in  series  (4)  we  obtain 

^tr  _  I  ^^^  ^  \ — A  _j.  \ L  -I-  i — ^  +  •  •  • 


2!         3! 
'I'      ^"^^!     6!"*"         > 

V     3!    55    7!        / 


A  comparison  with  series  (2)  and  (3)  shows  that  these  two 
series  are  cos;tr  and  sin;r  respectively;  hence  the  important 
formula  due  to  Euler —  « 

e«  =  cos  cc  +  i  sin  x,  (20) 

This  enables  us  to  calculate  e^  from  sin;ir  and  cos;ir  when 
tx  is  a  pure  imaginary  ;  that  is,  when  x  is  real. 

To  find  sim'x  and  cos  t'x  replace  :r  in  (20)  by  t'x;  we  obtain 

tf~*  =  cos  ix  +  /  sin  tx.  (21) 

Again  replacing  ;*:  by  —  t'x  in  (20),  we  obtain 

^  =  cos  ix  —  i  sin  ix.  (as) 

The  sum  and  difference  of  (21)  and  (22)  give 

COS  IX  =  — ■ =  cosh  X,  (23) 

sin  ix=  — — — — -  =  i  sinh  x,  (24) 

2 

If  we  compute  the  value  of  ^  by  the  aid  of  series  (4)  for 
a  succession  of  values  of  x,  we  find  that  sinh;tr  and  cosh;i: 
are  represented  by  the  curves  on  page  ^6. 

The  system  of  formulas  belonging  to  the  hyperbolic  func- 
tions is  obtained  from  those  of  the  trigonometric  functions 
by  using  (23)  and  (24).  This  shows  that  for  every  formula 
in  analytic  trigonometry  there  exists  a  corresponding  for- 
mula in  hyperbolic  trigonometry  which  we  get  by  this  sub- 


76 


PLANE   TRIGONOMETRY 


stitution.  In  the  examples  which  follow,  this  method  is 
used  to  obtain  important  formulas  in  hyperbolic  trigonome- 
try. 

Replacing  xhy  —ix  in  (23)  and  (24),  we  get 


008  07  = 


sin  07  = 


2 
3i      ' 


(25) 
(26) 


rhich  are  formulas  frequently  used. 


Example. — sinh  {x  -f /)  =  —  /  sin  i{x  •\-y\ 

=  —  z  [sin  ix  cos  //  -|-  cos  ix  sin  ty\, 

=  —  /  [/  sinh  X  cosh  j-f-  /  cosh  x  sinh/], 

=  sinh  jr  cosh _y-l- cosher  sinh_y. 

Example. — sinh  jr-f- sinh^  =  —  /(sin  ix  -}-  sin  iy), 

=  —  /  2  sin  ^  i{x-\-y)  cos  ^  i{x — y\ 
=  2  sinh  i  (x-^y)  cosh  ^  ix—y). 


sinh  oc 


cosh  X 


HYPERBOLIC  FUNCTIONS  77 

EXERCISES 
76.  (i.)'Prove  sinho=o,    cosho=i. 
(2 .)  Prove  si  n  h  ^i  =  /,     cosh  ^/ =o. 
(3.)  Prove  sinh7r/  =  o,     cosh7r/= — i. 

Prove  that 

(4.)  sin  (—  z'x)  —  —  sin  ix. 

(5 .)  cos  ( — t'x)  =  cos  zx. 

(6.)  sinh( — x)  =  —  sinhjr. 

(7.)  cosh  (—x)  =  cosh  ;r. 

Remark. — The  hyperbolic  tangent,  cotangent,  secant,  and  cosecant 

are  defined  by 

^     ,  sinh;jr  ,  cosh;ir 

tanh;r  = ; — .  coth;tr  =  ^ — . 

coshjr  sinh^ 

sech  X  =  — ; — f  csch  x  = 


cosh;r  sinh;ir 

Prove  ♦•hat 

(8.)  tan  {ix)  =  /  tanh  x. 

(9.)  coth  {—x)  =  —  coth  X. 

(10.)  sech  {—x)  =  sech  x. 

(II.)  cosh*;ir  — sinh'jr=:l. 

(12.)  sech'^4-tanh'jr=  I. 

(1 3.)  coth"^  —  csch*;i:  =  I . 

(14.)  sinh  {x  — y)  =  sinh  ;r  cosh j  —  cosh  x  shih,^. 

(15.)  cosh(;r — j)  =  cosh;r  cosh^  — sinh;r  sinh^. 

(16.)  coshi;r=y/i±^i5£. 

(17.)  sinh«  —  sinhz/  =  2  cosh  ^(«-|-t/)  sinh  ^(« — v). 

(18.)  cosh  « -j- cosh  2/ =  2  cosh  ^(?^-f-^)  cosh  ^(«—z/). 

(19.)  cosh  u  —  cosh  v  =  2  smh.^{u-{-v)  sinh  ^  (« — «/), 


CHAPTER  VII 
MISCELLANEOUS     EXERCISES, 

RELATION   OF  FUNCTIONS 

77'  Prove  the  following: 

(I.)  cos;r  =  sinjr  cot^. 

(2.)  cscjr  tan  ;r  =  seer. 

(3.)  (tan  ;ir -1- cot x)  sin.r  cos^  =  I. 

(4.)  (sec^  —  tan  J/)  {^secy  -\-  lainy)=l. 

(5.)  {cscz  —  col  z)  {CSC  z -{- col  2)  =  I . 

(6.)  cos''j  +  (tan/  — cot/)  sinj/  cosy  =  sin'j/. 

(7.)  cos*.r  — sin*.i--f  I  =2  cos^x. 

(8.)  (sin J  —  cos  )')'■  =  1  —  2  s'lny  cosy. 

(9.)  sin^r-f-cos'.r  =  (sin.r-}-cosx)  (i  — sin.r  cosx). 

,  cot  .r  4- tan  v 

(10.)  ; ^=:  cot.r  tan  y. 

tan  ^4- col)' 

(II.)  cos" J/  —  sin^/  =  2  cos''_y  —  I. 

(12.)  I  —  tan*  r  =  2  sec'-  x  —  sec*;r. 

cos  X 

(13.)  -: ;Tr-  =  tan.r. 

"^     sin.r  cot  .r 

(14.)  sec^j  csc^j'  =  tan-j-|-cot''/-4-2. 
(15.)  cot  J  —  cscy  sec_y  (i  —  2  sin^j/)  =  tanj'. 
—  cos  5' 


(16.)  (  .'     -cot.-)"  =  ^ 
\sin2-  /       I 

(17.) 


^COS^" 

sec  J/         tan  r  —  s\ny 


i+cos/  s\xvy 

(18.)  I  +  ^^-^ — ^  =  (sinx4-cos;jr)'. 
sec  X 

(IQ.)  ; sin^i=(cos-r  — sin;r)  (i  -)-sin  x  cos;r). 

'     sec  .r 

(20.)  (sin-r  cos/-f"Cos;r  sin ^)*-)- (cos  x  cosy  —  sin  x  sin /)"  =  !. 


MISCELLANEOUS  EXERCISES  79 

(21.)  [a  co%x  —  b  sin;ir)'  +  («  %\nx-^b  cos;ir)'  =  a'+^. 


(cos'/  — sin'^')"  (I  —  tan"/)' 

Find  an  angle  not  greater  than  90°  which  satisfies  each  of  the  fol- 
lowing equations : 
(23.)  4  cos  jr  =  3  sec  ^. 
(24.)  sin/ =  CSC/  — f. 
(25.)  \/2  sinjr  — tan;i;=0. 

(26.)  2  cos^  —  Vs  cot;ir=0. 

(27.)  tan/+ cot/— 2  =  o. 

(28.)  2  sin";'— 2  =  —V2  cos/. 

(29.)  3  tan"  jir  —  I  =  4  sin*  ^. 

(30.)  COS''jr-|-2  sin'';tr— f  sin:r  =  0. 

(31.)  csc;jr  =  f  tanjr. 

(32.)  sec  jr-l-  tan  x  =  -±.  ^3. 

(33.)  tan;r-|-2  ^3  cos;i;  =  a 

(34.)  3sin;ir  —  2cos'';f=o. 

Express  the  following  in  terms  of  the  functions  of  angles  less 
than  45°: 
(35.)  sin  92° 
(36.)  cos  127®. 
(37.)  tan  330°. 
(38.)  cot  350°. 
(39.)  sin  265°. 
(40.)  tan  171®. 

{41.)  Given  sin;r=^  and  x  in  quadrant  II;  find  all  the  other 
functions  of  x. 

(42.)  Given  cos-r  =  —  f  and  x  in  quadrant  III;  find  all  the  other 
functions  of  x. 

C43.)  Given  tanjr  =  :i  and  x  in  quadrant  III;  find  all  the  other 
functions  of  x. 

(44.)  Given  cotx  =  --|  and  x  in  quadrant  IV;  find  all  the  other 
functions  of  x. 


80  PLANE    TRIGONOMETRY 

In  what  quadrants  must  the  angles  lie  which  satisfy  each  of  the 
following  equations : 

(45.)  sin  X  cos  ;r  =  —  \'\pi- 
(46.)  sec  X  tan  jr  =  2  -y/3. 

{47.)  tan_>'4- -y/20  cos>'  =  a 
(48.)  cos  X  cot  ^  =  f . 

Find  all  the  values  of  y  less  than  360°  which  will  satisfy  tne  fol- 
lowing equations : 

(49.)  tanj-f-2  sinj  =  o. 

(50.)  (I  -f-  tan  x)  (I  —  2  sin  x)  =0. 

(51.)  sinjT  cos;r  (i-f-2  cos-r)=a 

Prove  the  following: 
(52.)  cos  780°  = -J, 

(53.)  sin  1 4850  =  iV^. 

(54.)  cos255o°  =  ^-/3- 

(55.)  sin(— 3000°)  =  — cos  30P. 

(56.)  cos  1 300°= — cos  40*^. 

(57.)  Find  the  value  of  a  sin  90° -f^  tan  cP-^a  cos  180P. 

(58.)  Find  the  value  of  a  sin  30°  +  (5  tan  45° -f  «  cos  60° -f-*^  tan  135°. 

(59.)  Find  the  value  of  {a  —  b)  tan  225° -f'^  cos  180°  —  a  sin  270°. 

(60.)  Find  the  value  of  (a  sin  45°+ <J  cos  45°)  (a  sin  i35°  +  iJsin  225°). 

RIGHT  TRIANGLES 

'TIS.  In  the  following  problems  the  planes  on  which  distances  are  measured 
are  understood  to  be  horizontal  unless  otlierwise  stated. 

(I.)  The  angle  of  elevation  of  the  top  of  the  tower  from  a  point 
1 121  ft.  from  its  base  is  observed  to  be  15°  17';  find  the  height  of 
the  tower. 

(2.)  A  tree,  'J^  ft.  high,  stands  on  the  bank  of  a  river;  at  a  point  on 
the  other  bank  just  opposite  the  tree  the  angle  of  elevation  of  the 
top  of  the  tree  is  found  to  be  5°  17' 37".  Find  the  breadth  of  the 
river. 


MISCELLANEOUS  EXERCISES  8l 

(3.)  What  angle  will  a  ladder  42  ft.  long  make  with  the  ground  if  its 
foot  is  25  ft.  from  the  base  of  the  building  against  which  it  is  placed  } 

(4.)  When  tne  altitude  of  the  sun  is  33°  22',  what  is  the  height  of  a 
tree  which  casts  a  shadow  75  ft.? 

(5.)  Two  towns  are  3  miles  apart.  The  angle  of  depression  of  one, 
from  a  balloon  directly  above  the  other,  is  observed  to  be  8°  15'. 
How  high  is  the  balloon  ? 

(6.)  From  a  point  197  ft.  from  the  base  of  a  tower  the  angle  of  ele- 
vation was  found  to  be  46°  45'  54" ;  find  the  height  of  the  tower. 

(7.)  A  man  5  ft.  10  in.  high  stands  at  a  distance  of  4  ft.  7  in.  from 
a  lamp-pq^t,  and  casts  a  shadow  18  ft.  long;  find  the  height  of  the 
lamp- post. 

(8.)  The  shadow  of  a  building  101.3  ft.  high  is  found  to  be  131. 5 
ft.  long;  find  the  elevation  of  the  sun  at  that  time. 

(9.)  A  rope  112  ft.  long  is  attached  to  the  top  of  a  building  and 
reaches  the  ground,  making  an  angle  of  ^^°  20'  with  the  ground; 
find  the  height  of  the  building. 

(10.)  A  house  is  130  ft.  above  the  water,  on  the  banks  of  a  river; 
from  a  point  just  opposite  on  the  other  bank  the  angle  of  elevation 
of  the  house  is  14°  30'  21".     Find  the  width  of  the  river. 

(II.)  From  the  top  of  a  headland,  1217.8  ft.  above  the  level  of  the 
sea,  the  angle  of  depression  of  a  dock  was  observed  to  be  10°  9'  13"  •/ 
find  the  distance  from  the  foot  of  the  headland  to  the  dock. 

(12.)  1 121. 5  ft.  from  the  base  of  a  tower  its  angle  of  elevation  is 
found  to  be  11°  3'  5";  find  the  height  of  the  tower. 

(13.)  One  bank  of  a  river  is  94.73  ft.  vertically  above  the  water,  and 
subtends  an  angle  of  10°  54'  13"  from  a  point  directly  opposite  at  the 
water's  edge;  find  the  width  of  the  river. 

(14.)  The  shadow  of  a  vertical  cliff  113  ft.  high  just  reaches  a  boat 
on  the  sea  93  ft.  from  its  base ;  find  the  altitude  of  the  sun. 

(15.)  A  rope,  38  ft.  long,  just  reached  the  ground  when  fastened  to 
the  top  of  a  tree  29  ft.  high.  What  angle  does  it  make  with  the 
ground  ? 

(16.)  A  tree  is  broken  by  the  wind.  Its  top  strikes  the  ground  1 5 
ft.  from  the  foot  of  the  tree,  and  makes  an  angle  of  42°  28'  with  the 
ground.     Find  the  height  of  the  tree  before  it  was  broken. 


83  PLANE  TRIGONOMETRY 

(17.)  The  pole  of  a  circular  tent  is  18  ft.  high,  and  the  ropes  reach- 
ing from  its  top  to  stakes  in  the  ground  are  37  ft.  long;  find  the 
distance  from  the  foot  of  the  pole  to  one  of  the  stakes,  and  the  angle 
between  the  ground  and  the  ropes. 

(18.)  A  ship  is  sailing  southwest  at  the  rate  of  8  miles  an  hour. 
■At  what  rate  is  it  moving  south  ? 

(19.)  A  building  is  121  ft.  high.  From  a  point  directly  across  the 
street  its  angle  of  elevation  is  65°  3'.     Find  the  width  of  the  street. 

(20.)  From  the  top  of  a  building  52  ft.  high  the  angle  of  elevation 
of  another  building  112  ft.  high  is  30°  12'.  How  far  are  the  buildings 
kpart? 

(21.)  A  window  in  a  house  is  24  ft.  from  the  ground.  What  is  the 
inclination  of  a  ladder  placed  8  ft.  from  the  side  of  the  building  and 
reaching  the  window  ? 

(22.)  Given  that  the  sun's  distance  from  the  earth  is  92,000,000 
miles,  and  its  apparent  semidiameter  is  16'  2" ;  find  its  diameter. 

(23.)  Given  that  the  radius  of  the  earth  is  3963  miles,  and  that  it 
subtends  an  angle  of  57'  2"  at  the  moon;  find  the  distance  of  the 
moon  from  the  earth. 

(24.)  Given  that  when  the  moon's  distance  from  the  earth  is  238885 
miles,  its  apparent  semidiameter  is  15'  34" ;  find  its  diameter  in  miles. 

(25.)  Given  that  the  radius  of  the  earth  is  3963  miles,  and  that  it 
subtends  an  angle  of  9"  at  the  sun;  find  the  distance  of  the  sun 
from  the  earth. 

(26.)  A  light-house  is  57  ft.  high  ;  the  angles  of  elevation  of  the  top 
and  bottom  of  it,  as  seen  from  a  ship,  are  5°  3'  20"  and  4°  28'  8".  Find 
the  distance  of  its  base  above  the  sea-level. 

(27.)  At  a  certain  point  the  angle  of  elevation  of  a  tower  was  ob- 
served to  be  53°  51'  16",  and  at  a  point  302  ft.  farther  away  in  the 
same  straight  line  it  was  9°  52'  10";  find  the  height  of  the  tower. 

(28.)  A  tree  stands  at  a  distance  from  a  straight  road  and  between 
two  mile-stones.  At  one  mile-stone  the  line  to  the  tree  is  observed 
to  make  an  angle  of  25°  15'  with  the  road,  and  at  the  other  an  angle 
of  45°  17'.     Find  the  distance  of  the  tree  from  the  road. 

(29.)  From  the  top  of  a  light-house,  225  ft.  above  the  level  of  the 
sea,  the  angles  of  depression  of  two  ships  are  i  7°  2  i'  50"  and  13°  50'  22", 


MISCELLANEO  US  EXERCISES  83 

and  the  line  joining  the  ships  passes  directly  beneath  the  light-house  : 
find  the  distance  between  the  two  ships. 

ISOSCELES  TRIANGLES  AND   REGULAR   POLYGONS 

79,  (I.)  The  area  of  a  regular  dodecagon  is  37.52  ft. ;  find  its 
apothem. 

(2.)  The  perimeter  of  a  regular  polygon  of  1 1  sides  is  23.47  ft. ;  find 
the  radius  of  the  circumscribing  circle. 

(3.)  A  regular  decagon  is  circumscribed  about  a  circle  whose  radius 
is  3.147  ft. ;  find  its  perimeter. 

(4.)  The  side  of  a  regular  decagon  is  23.41  ft. ;  find  the  radius  of 
the  inscribed  circle. 

(5.)  The  perimeter  of  an  equilateral  triangle  is  17.2  ft.;  find  the 
area  of  the  inscribed  circle. 

(6.)  The  area  of  a  regular  octagon  is  2478  sq.  in. ;  find  its  pe- 
rimeter. 

(7.)  The  area  of  a  regular  pentagon  is  32.57  sq.  ft. ;  find  the  radius 
of  the  inscribed  circle. 

(8.)  The  angle  between  the  legs  of  a  pair  of  dividers  is  43°  and  the 
legs  are  7  in.  long;  find  the  distance  between  the  points. 

(9.)  A  building  is  37.54  ft.  wide,  and  the  slope  of  the  roof  is  43"  36'; 
find  the  length  of  the  rafters. 

(10.)  The  radius  of  a  circle  is  12732,  and  the  length  of  a  chord  is 
18321 ;  find  the  angle  the  chord  subtends  at  the  centre. 

(II.)  If  the  radius  of  a  circle  is  taken  as  unity,  what  is  the  length 
of  a  chord  which  subtends  an  angle  of  ']']'^  17'  40"? 

(12.)  What  angle  at  the  centre  of  a  circle  does  a  chord  which  is  ^ 
of  the  radius  subtend  ? 

(13.)  What  is  the  radius  of  a  circle  if  a  chord  11223  ft.  subtends  an 
angle  of  50°  50'  52"  ? 

(i^.;  Two  light-houses  at  the  mouth  of  a  harbor  are  each  2  miles 
from  the  wharf.  A  person  on  the  wharf  finds  the  angle  between  the 
lines  to  the  light-houses  to  be  17°  33'.  Find  the  distance  between  the 
two  light-houses. 

(15.)  The  side  of  a  regular  pentagon  is  2;  find  the  radius  of  the 
inscribed  circle. 


84  PLANE    TRIGONOMETRY 

(1 6.)  The  perimeter  of  a  regular  heptagon  inscribed  in  a  circle  is 
12  ;  find  the  radius  of  the  circle. 

(17.)  The  radius  of  a  circle  inscribed  in  an  octagon  is  3;  find  the 
perimeter  of  the  octagon. 

(18.)  A  regular  polygon  of  9  sides  is  inscribed  in  a  circle  of  unit 
radius;  find  the  radius  of  the  inscribed  circle. 

(19.)  Find  the  perimeter  of  a  regular  decagon  circumscribed  about 
a  unit  circle. 

V 

(20.)  Find  the  area  of  a  regular  hexagon  circumscribed  about  a 
unit  circle. 

(21.)  Find  the  perimeter  of  a  polygon  of  11  sides  inscribed  in  a 
unit  circle. 

(22.)  The  perimeter  of  a  dodecagon  is  30;  find  its  area. 

(23.)  The  area  of  a  regular  polygon  of  1 1  sides  is  18;  find  its  pe- 
rimeter. 

TRIGONOMETRIC   IDENTITIES  AND   EQUATIONS 
80.  Prove  the  following: 


(I.)  sin  ^/±cos  |/  = -\/i  itsin/. 

,    ,  cos  X  —  cos  y  ■,,,•.         1  /  s 

(2.)  -  ,         -^  =  —tan  ^  (x  -}-_y)  tan  ^  {x  —y). 

cos  x  + cos/  ^         -^  -^ 

,    .   sin  2x  4-  sin^r 

(3.) ; ^—  =  tan  2,x. 

cos  2x  -\-  cos  ^x 


)  cos- J  tan'^j' -f-  sin^y  col^y  =  I. 

.    CO?,  (x  4- y -\- s) 

)  —. —     .         . —  =  cot  X  cot  y  cot  z  —  cot  x  —  cot  y  —  cot  z. 
sin.r  ?,\ny  sm  s 

)  cos'^  (.r  —y)  —  sin''  {x-\-y)  =  cos  2x  cos  2_y. 

,    sin, r 4- sin  y  ,  ^  ^ 

cos  x  —  cos_y 


^o^  cos.r  — sec;t-  2  ,      /       si  .x 

(8.)  =  4  cos^  4  X (cos''  hx—l). 

seer  8     \         s 

sin  2x 

(9.)  cot;r  = ■• 

^^  I  —  cos  2X 

,         I  —  cos  2y 

(10.)  tanV  =  — ; ^  • 

^  -^       I  -[-  cos  2y 

(II.)  cot  X  —  tan  .r  =  2  cot  2x. 


(12 

(13 

(H 
(15 

(16 

(17. 
(18, 

(19 
(20. 

(21. 

(22. 

(23. 

(24. 

(25. 
(26. 

(27. 

(28. 

(29 
(30- 

(3^ 

(32 
(33 
(34 

(35. 
(36. 


MISCELLANEOUS  EXERCISES  85 

tan  ^  -tr  -j-  2  sin'  ^  ;r  cot  x  =  sin  jr. 

tan  X  ±  tan  y 

; —=  ±  sin  X  sec  jr  tan  v. 

cot  X  ±  cot^  -^ 

sin  X  —  2  sin'  ;tr  =  sin  ;r  cos  2x. 

4  sin  J/  sin  (60° — y)  sin  (60°  -^-y)  =  sin  s^j'. 

sin  y  (I— tan' y)/  i  .  i  \        . 

— " \ \ ■■ i — ■■ — )  =  sm  2  y. 

sec' J/  xcosj — sin^      cosj-f-sin//  "^ 

I  -f-  tan/  tan  ^  j  =rsec/. 

sin  4^  =  4  sin  x  cos'  x  —  4  cos  x  sin'  jr. 

2 

sec  2-r  +  tan  2;tr  +  i  = • 

I  —  tan  X 

tan  50°+ cot  50°  =  2  sec  10°. 

cos  (Jf  +  45°)  +  sin  (;r  —  45°)  =  o. 

tanjr 


I  —  cot  2;r  tan  x 


:=Sin  2X. 


—  C06  2;r 


(I  —  tan'  x')  sin  x  cos  x  =  cos  2x  \J — 

cos  y -I- sin  V 

— ^-^  =tan  2y+sec2y. 

cos_y  — sinj  -^  "^ 

sin  {x-\-y)  cos-r — cos(^-j-/)  sinx  — sinj^. 

cos  (x — y)  sin_y  -|-  sin  {x  —y)  cosj  =  sin  x. 

sin  (x—y)        s\n{y — z)       sin  {z~x)  _ 

cos-r  cos  J/     cos/  cos  2"     cos  2"  cos-r 

sin,r4-sin  2;ir 
=  cot  A  X. 

cos  X  —  cos  2X 

2  sin'  X  sin'/+  2  cos'  x  cos'/  =  i  4-  cos  2x  cos  2jl^ 
sin  60°+  sin  30°  =  2  sin  45°  cos  1 5°. 
tan  (^— /)  +  tan/ 
I  —  tan  (x—.y)  tany 
2 


—  tan  X. 


i+cot'^/. 


sin/  tan  ^/ 
sin  4.r-(-sin  2-r  =  2  sin  ■^x  cos  jr. 
sinjr  +  sin/  _cosjf  +  cos/ 
cos-r  —  cos/       sin/  —  sinjr 

sin75°=         7^    • 
2-v/ 2 

2 tan  2/  =  tan(45°+/)  — tan(45°— /). 


86  PLANE    TRIGONOMETRY 

tan  2  ;»r  4- tan  ^      sin3Jr 


(37 

(38. 

(39. 
(40. 

(41. 

(42. 

(43. 
(44 

(45. 
(46 

(47. 
(48, 

(49 
(50. 

(51. 

(52- 
(53. 
(54 

(55- 

(56. 

(57. 
(58 

(59. 

(60 
(61 


tan  2^—;  tan  j:       sin;r 

3  tan  y  — tan' y 

tan3j  =  -2 ^— — 5-^. 

1—3  tan*/ 

sin 60° 4-  sin  20°  =  2  sin 40°  cos2oP. 

sin40°  —  sin  io°  =  2  00325°  sin  15". 

cos  24: — cos4-r  =  2  sin3;r  sinjr. 

tan  150  =  2 — Vs- 

(■y/i  +  sinx  — -v/i  — sin^)*  =  4  sin'^jr. 

(-/i  +sin^-f -\/i  — sin  jr)''  =  4  cos'^X 

sin(2^  +  y)  ,     ,     .       sin_y 
^^^ — —~  —  2  cos(;r4-j)  =  -r-^' 


sm^  sinjr 

sin4;r 

^—  =  2  cos  2;r. 

sin2;ir 

sin  50°  — sin  70° -f  sin  10"  =  0. 

TT  TT  .       5^     •       W 

COS cos  -  =  2  sin  ^—  sin  — 

32  12       12 

I  —  tan'^(450— ,r) 

-1 ,^-7^ —  ;  =  sm2r. 

I  4- tan"  (4 5°  —  ^) 

si 

COS/i" -fuus  1^-         V    i 


3in75°  — sin  15°  _      /f 
;os  7  5°  4-  cos  1 5°  ~~  V  ^  * 


tan»4;r(i  +  cot' I  ^)' = -^T- 

tan  75°  =  2  4- V^. 

sin  yc  4-  sin  5^  =  2  sin  4-r  cos  jr. 

cos  5  jr  4-  cos  9jr  =  2  cos  7-ir  COS  XT. 

sin  15°=  ■'^-~-' 

2V2 

sin3.r  — sinjT 


=  tan  X. 


cos  3x  +  cos  X 

sin  5_y  =  5  sinj/  —  20  sin'j/-|-  16  sin^. 

cos  5/  =  5  cos/  —  20  cos'/  4- 16  cos^. 

4  tan X (I — tan'^jr) 

sin4f= — j— — j--^ 

(1+ tan. IT 

cos(45<5-|-.r)-}-cos(45°  — ;t)=:  •v/2  cos^f. 

cos  yc  4-  cos  5-r  4-  cos  ^x  +  cos  1 5-f  =  4  cos  \x  cos  5  x  cosSx". 


MISCELLANEOUS  EXERCISES  87 

(62.)  sin''^;r(cot^;r — i)*  =  i— sin;r. 

,,   ^    3sin^  —  sin3;r 

(630 i ^-  =  tan»^. 

cos  3-^4-3  cos  X 

(64,)  sin  ;r  ( I  -f-  tan  x)  -^  cos ^ (i  +  cot x)  =  esc x  ■\-  sec j:. 

,-    ,  cos'^  —  sin'jr      24-sin2;ir 

(65.)    : =— n 

cos^  —  sin^  2 

(66.)  cosj/  -j-  cos  ( 1 20  — y)  -\-  cos  ( 1 20  \-y)  =  o. 
,^    .   sin3jr 

(67.)       .  =2  COS2X+  I. 

sin^r 

(cosj—  cos  3/)(sin  8j-|-  sin  2y)  _  ^ 

(sin  y  —  sin^)(cos4y  —  cos 6/) 


,,    .    /    sin;ir    \^      i 


-f-cos;r/       i-}-cosjr 

.   sinrr     cos3;r 

(70.)  -^-^  — ^  =  2. 

sinj:        cos^r 

^   I -I- sin  ;r  4- cos X 

(7I-)  — F^ =cot4x 

\-\-%\\\x  —  cos;ir 

,     .    sin(4,r  — 2y)-|-sin(4v  — 2X)      ^      ,     ,     ^ 
(72.)  ^-^ ^^ -^ '.=xjaxi{x-\'y). 

cos  (4^  —  2j) -f- cos  (4,^  —  2jr) 

^    sin  x  + sin  3jr  + sin  5jr  4- sin  7Jr 

(73.)  7- ^-—T T  =tan4T. 

"*^     cos  ;r  + cos  3-1:4- cos  54: 4- cos  7  JT 

If  A,  B,  and  C  are  the  angles  of  a  triangle,  prove  the  following-; 
(74.)  sin  2^  4-  sin  2i5  -|-  sin  2C  =  4  sin  ^  sin  5  sin  C. 
(75.)  sin  2A  4-  sin  2;9  — sin  2C  =  4  cos^  cos^  sinC 
(76.)  sin''/i  +  sin-i>'  +  sin''C=  2  +  2  cos^  cos 5  cosC 
{^"j^  tan  A  •\-  tan  B  4-  tan  C  =  tan  A  tan  B  tan  C. 

Solve  the  following  equations  for  values  of  x  less  than  360°. 

(78.)  cos  2^ -f  cos .r  =  —  I. 

(79.)  sin.r4-sin7.r  =  sin4-r. 

(80.)  cos.r  —  sin2jr  —  cos3Jr  =  o. 

(81.)  cos^  — sin3;r  —  cos2-ir  =  o. 

(82.)  sin4;r  —  2  sin  2 jt  1=0. 

(83.)  sin  2^  —  COS2X  — sinA-4-cos4r  =  o. 

(84.)  sin (60°  —  ;r)  —  si n  (60° -\-x)  =  -\-^  y 3^ 

(85.)  sin  (30°  +  x)  —  cos  (60°  -\-x)  =  —^^^^ 


8S  PLANE    TRIGOJSIUMETRY 

(86.)  csc;r=  I  4-cot;ir. 

(87.)  cos2jr  =  cos'4r. 

(88.)  2  sin^=sin2y. 

(89.)  sin3/  +  sin2y-|-sinj/  =  a 

(90.)  sin^r  -|-  5  cosV  =  3. 

(91.)  tan(45°— x)-|-cot(45*-r)s=4. 


OBLIQUE  TRIANGLES 

81,  (I.)  It  is  required  to  find  the  distance  between  two  points,  A 
and  B,  on  opposite  sides  of  a  river.  A  line, /4C  and  the  angles  BAC 
and  ACB  are  measured  and  found  to  be  2483  ft.,  61°  25',  and  52°  \^' 
respectively. 

(2.)  A  straight  road  leads  from  a  town  .^  to  a  town  B,  12  miles 
distant ;  another  road,  making  an  angle  of  'J^°  with  the  first,  goes  from 
/4  to  a  town  C,  7  miles  distant.     How  far  are  the  towns  B  and  C  apart  ? 

(3.)  In  order  to  determine  the  distance  of  a  fort.  A,  from  a  battery, 
B,  a  line,  BC,  one-half  mile  long,  is  measured,  and  the  angles  ABC 
and  ACB  are  observed  to  be  75°  18'  and  78°  21'  respectively.  Find 
the  distance /^^. 

(4.)  Two  houses,^  and  B,  are  1728  ft.  apart.  Find  the  distance  of 
a  third  house,  C,  from  A  if  BAC—M°  51'  and  ABC=S7°  23'. 

(5.)  In  order  to  determine  the  distance  of  a  bluff,  A,  from  a  house, 
B,  in  a  plane,  a  line,  BC,  was  measured  and  found  to  be  1281  yards, 
also  the  angles  ABC  and  BCA  65°  31'  and  70°  2'  respectively.  Find 
the  distance  AB. 

(6.)  Two  towns,  3  miles  apart,  are  on  opposite  sides  of  a  balloon. 
The  angles  of  elevation  of  the  balloon  are  found  to  be  13°  19'  and 
20°  3'.     Find  the  distance  of  the  balloon  from  the  nearer  town. 

(7.)  It  is  required  to  find  the  distance  between  two  posts,  y?  and  B, 
which  are  separated  by  a  swamp.  A  point  C  is  1272.5  ft.  from  A,  and 
2012.4  ft.  from  B.     The  angle  ^C^  is  41°  9'  11". 

(8.)  Two  stakes,  A  and  B,  are  on  opposite  sides  of  a  stream  ,■  a 
third  point,  C,  is  so  situated  that  the  distances  AC  and  BC  can  be 
found,  and  are  431.27  yards  and  601.72  yards  respectively.  The  angle 
ACB  is  39°  53'  13".     Find  the  distance  between  the  stakes  A  and  B. 


MISCELLANEOUS   EXERCISES  89 

(9.)  Two  light-houses,  A  and  B,  are  1 1  miles  apart.  A  ship,  C,  is 
observed  from  them  to  make  the  angles  BAC='^\°  13'  31'  and  ABC 
=21°  46'  8".     Find  the  distance  of  the  ship  from  A. 

(10.)  Two  islands,  A  and  B,  are  6103  ft.  apart.  Find  the  distance 
from  A  too.  ship,  C,  if  the  angle  ABC  is  37°  25'  and  BAC  is  40°  32'. 

(II.)  In  ascending  a  cliff  towards  a  light-house  at  its  summit,  the 
light-house  subtends  at  one  point  an  angle  of  21°  22'.  At  a  point 
55  ft.  farther  up  it  subtends  an  angle  of  40P  27'.  If  the  light-house 
is  58  ft.  high,  how  far  is  this  last  point  from  its  foot? 

(12.)  The  distances  of  two  islands  from  a  buoy  are  3  and  4  miles 
respectively.  The  islands  are  2  miles  apart.  Find  the  angle  sub- 
tended by  the  islands  at  the  buoy. 

(13.)  The  sides  of  a  triangle  are  151.45,  191.32,  and  250.91.  Find 
the  length  of  the  perpendicular  from  the  largest  angle  upon  the 
opposite  side. 

(14.)  A  tree  stands  on  a  hill,  and  the  angle  between  the  slope  of  the 
hill  and  the  tree  is  110^23'.  At  a  point  85.6  ft.  down  the  hill  the 
tree  subtends  an  angle  of  22°  22'.     Find  the  height  of  the  tree. 

(15.)  A  light-house  54  ft.  high  is  built  upon  a  rock.  From  the  top 
of  the  light-house  the  angle  of  depression  of  a  boat  is  19°  10',  and 
from  its  base  the  angle  of  depression  of  the  boat  is  12°  22'.  Find  the 
height  of  the  rock  on  which  the  light-house  stands. 

(16.)  Three  towns.  A,  B,  and  C,  are  connected  by  straight  roads. 
AB^iit  miles,  BC-=  5  miles,  and  AC=^^  miles.  Find  the  angle  made 
by  the  roads  AB  and  BC. 

(17.)  Two  buoys,  A  and  B,  are  one-half  mile  apart.  Find  the  dis- 
tance from  ^  to  a  point  C  on  the  shore  if  the  angles  ABC  and  BAC 
are  ^^'^  7'  and  67°  17'  respectively. 

(18.)  The  top  of  a  tower  is  175  ft.  above  the  level  of  a  bay.  From 
its  top  the  angles  of  depression  of  the  shores  of  the  bay  in  a  certain 
direction  are  57°  16'  and  15°  2'.     Find  the  distance  across  the  bay. 

(19.)  The  lengths  of  two  sides  of  a  triangle  are  ■y/T.  and  ■y/'^.  The 
angle  between  them  is  45°.     Find  the  remaining  side. 

(20.)  The  sides  of  a  parallelogram  are  172.43  and  101.31,  and  the 
angle  included  by  them  is  61°  16'.     Find  the  two  diagonals. 

(21.)  A  tree  41  ft.  high  stands  at  the  top  of  a  hill  which  slopes 


9©  PLANE  TRIGONOMETRY 

io°  12'  to  the  horizontal.  At  a  certain  point  down  the  hill  the  tree 
subtends  an  angle  of  28°  29'.  Find  the  distance  from  this  point  to 
the  foot  of  the  tree. 

(22.)  A  plane  is  inclined  to  the  horizontal  at  an  angle  of  7°  33'. 
At  a  certain  point  on  the  plane  a  flag-pole  subtends  an  angle  20°  3', 
and  at  a  point  50  ft.  nearer  the  pole  an  angle  of  40°  35'.  Find  the 
height  of  the  pole. 

(23.)  The  angle  of  elevation  of  an  inaccessible  tower,  situated  in 
a  plane,  is  53°  19'.  At  a  point  227  ft.  farther  from  the  tower  the 
angle  of  elevation  is  22°  41'.     Find  the  height  of  the  tower. 

(24.)  A  house  stands  on  a  hill  which  slopes  12°  18' to  the  horizon- 
tal. 75  ft.  from  the  house  down  the  hill  the  house  subtends  an 
angle  of  32°  5'.     Find  the  height  of  the  house. 

(25.)  From  one  bank  of  a  river  the  angle  of  elevation  of  a  tree  on 
the  opposite  bank  is  28°  31'.  Frc  a  point  139.4  ft.  farther  away  in  a 
direct  line  its  angle  of  elevation  is  i9°io'.  Find  the  width  of  the  river. 

(26.)  From  the  foot  of  a  hill  in  a  plane  the  angle  of  elevation  of 
the  top  of  the  hill  is  2 1°  7 '.  After  going  directly  away  2 1 1  ft.  farther, 
the  angle  of  elevation  is  18°  37'.     Find  the  height  of  the  hill. 

(27.)  A  monument  at  the  top  of  a  hill  is  153.2  ft.  high.  At  a  point 
321.4  ft.  down  the  hill  the  monument  subtends  an  angle  of  11°  13'. 
Find  the  distance  from  this  point  to  the  top  of  the  monument. 

(28.)  A  building  is  situated  on  the  top  of  a  hill  which  is  inclined 
10°  12'  to  the  horizontal.  At  a  certain  distance  up  the  hill  the  angle 
of  elevation  of  the  top  of  the  building  is  20°  55',  and  1 15.3  ft.  farther 
down  the  hill  the  angle  of  elevation  is  15°  10'.  Find  the  height  of 
the  building. 

(29.)  A  cloud,  C,  is  observed  from  two  points,  A  and  B,  2874  ft. 
apart,  the  line  AB  being  directly  beneath  the  cloud.  At  A^  the 
angle  of  elevation  of  the  cloud  is  77°  19',  and  the  angle  CAB  is 
51°  18'.  The  angle  ABC  is  found  to  be  60°  45'.  Find  the  height 
of  the  cloud  above  A. 

(30.)  Two  observers,  A  and  B,  are  on  a  straight  road,  675.4  ft. 
apart,  directly  beneath  a  balloon,  C.  The  angles  ABC  and  BAC 
are  34°  42'  and  41°  15'  respectively.  Find  the  distance  of  the  bal- 
loon from  the  first  observer. 


MISCELLANEOUS  EXERCISES  91 

(31.)  A  man  on  the  opposite  side  of  a  river  from  two  objects,  A 
and  B,  wishes  to  obtain  their  distance  apart.  He  measures  the  dis- 
tance CD  =  2,S7  ft.,  and  the  angles  ACB=2(f  33',  BCD  =  38°  52'.  ADB 
=  54°  10',  and  ADC=  34°  11'.     Find  the  distance  AB. 

(32.)  A  cliff  is  327  ft.  above  the  sea-level.  From  the  top  of  the 
cliff  the  angles  of  depression  of  two  ships  are  15°  11'  and  15°  13'. 
From  the  bottom  of  the  cliff  the  angle  subtended  by  the  ships  are 
122°  39'.     How  far  are  the  ships  apart  ? 

(33.)  A  man  standing  on  an  inclined  plane  112  ft.  from  the  bottom 
observed  the  angle  subtended  by  a  building  at  the  bottom  to  be  33^^ 
52'.  The  inclination  of  the  plane  to  the  horizontal  is  18°  51'.  Find 
the  height  of  the  building. 

(34)  Two  boats,  A  and  B,  are  451.35  ft.  apart.  The  angle  of  ele- 
vation of  the  top  of  a  light-house,  as  observed  from  A,  is  33°  17'. 
The  base  of  the  light-house,  C,  is  level  with  the  water;  the  angles 
ABC  and  CAB  are  12°  31'  and  137°  22'  respectively.  Find  the  height 
of  the  light-house. 

(35.)  From  a  window  directly  opposite  the  bottom  of'  a  steeple  the 
angle  of  elevation  of  the  top  of  the  steeple  is  29°  21'.  From  another 
window,  20  ft.  vertically  below  the  first,  the  angle  of  elevation  is  39°  3'. 
Find  the  height  of  the  steeple. 

(36.)  A  dock  is  I  mile  from  one  end  of  a  breakwater,  and  i|  miles 
from  the  other  end.  At  the  dock  the  breakwater  subtends  an  angle 
of  31°  11'.     Find  the  length  of  the  breakwater  in  feet. 

(37.)  A  straight  road  ascending  a  hill  is  1022  ft.  long.  The  hill 
rises  i  ft.  in  every  4.  A  tower  at  the  top  of  the  hill  subtends  an 
angle  of  7°  19'  at  the  bottom.     Find  the  height  of  the  tower. 

(38.)  A  tower,  192  ft.  high,  rises  vertically  from  ene  corner  of  a 
triangular  yard.  From  its  top  the  angles  of  depression  of  the  other 
corners  are  58^4'  and  17°  49'.  The  side  opposite  the  tower  subtends 
from  the  top  of  the  tower  an  angle  of  75°  15'.  Find  the  length  of 
this  side. 

(39.)  There  are  two  columns  left  standing  upright  in  a  certain  ruins ; 
the  one  is  66  ft.  above  the  plain,  and  the  other  48.  In  a  straight  line 
between  them  stands  an  ancient  statue,  the  head  of  which  is  100  ft. 
from  the  summit  of  the  higher,  and  84  ft.  from  the  top  of  the  lower 


9*  PLANE    TRIGONOMETRY 

column,  the  base  of  which  measures  just  74  ft.  to  the  centre  of  the 
figure's  base.  Required  the  distance  between  the  tops  of  the  two 
columns. 

(40.)  Two  sides  of  a  triangle  are  in  the  ratio  of  1 1  to  9,  and  the 

opposite  angles  have  the  ratio  of  3  to  i.     What  are  these  angles  ? 

(41*.)  The  diagonals  of  a  parallelogram  are  12432  and  8413,  and  the 
angle  between  them  is  78°  44';  find  its  area. 

(42.)  One  side  of  a  triangle  is  1012.6  and  the  two  adjacent 
angles  are  52°  21'  and  57°  32' ;  find  its  area. 

(43.)  Two  sides  of  a  triangle  are  218.12  and  123.72,  and  the  in- 
cluded angle  is  59°  10' ;  find  its  area. 

(44.)  Two  angles  of  a  triangle  are  35°  15'  and  47°  18',  and  the 
included  side  is  2104.7  j  ^"^  i^s  area. 

(45.)  The  three  sides  of  a  triangle  are  1.2371,  i.47i'^  and  2.0721 ; 
find  the  area. 

(46.)  Two  sides  of  a  triangle  are  168.12  and  179.21.  and  the  included 
angle  is  41°  14' ;  find  its  area. 

(47.)  The  three  sides  of  a  triangle  are  51  ft.,  48.12  ft.,  and  32.2  ft. ; 
find  the  area. 

(48.)  Two  sides  of  a  triangle  are  in. 18  and  121. 21,  and  the  included 
angle  is  27°  50' ;  find  its  area. 

(49.)  The  diagonals  of  a  parallelogram  are  37  and  51,  and  they  form 
an  angle  of  65°;  find  its  area. 

(50.)  If  the  diagonals  of  a  quadrilateral  are  34  and  56,  and  if  they 
intersect  at  an  angle  of  67°,  what  is  the  area? 


SPHERICAL  TRIGONOMETRY 

CHAPTER   VIII 

RIGHT  AND   QUADRANTAL  TRIANGLES 

RIGHT   TRIANGLES 

82,  Let  O  be  the  centre  of  a  sphere  of  unit  radius,  and 
ABC  a  right  spherical  triangle,  right  angled  at  A,  formed  by 
the  intersection  of  the  three  planes  ^(96",  A  OB,  and  BOC 


with  the  surface  of  the  sphere.  Suppose  the  planes  DAC" 
and  BEC  passed  through  the  points  A  and  B  respectively, 
and  perpendicular  to  the  line  OC.  The  plane  angles  DC" A 
and  BC E  each  measure  the  angle  C  of  the  spherical  tri- 
angle, and  the  sides  of  the  spherical  triangle  a,  b,  c  have  the 
same  numerical  measure  as  BOC,  AOC,  and  AOB  respec- 


94  SPHERICAL    TRIGONOMETRY 

tively,  then,  ^Z?  =  tanr,  BE  =  %\nc,  BC  =  s'in  a,  OC  =  cosa, 

OC"  =  cos d,  OE=cosc,  AC" =sin  b. 

In  the  two  similar  triangles  OEC  and  OAC", 

cos  c     cos  c      cos  a  ,  ,  . 

_  ■  = = 7 ,  or  cos ^  =  cos ^  cos r.  (i) 

OA  I         cos  if  ^  ^ 

In  the  triangle  BC'E, 

.    ^     BE  .    ^     sin  f  /  V 

sin  6  =  -;r7Tv,  or  sin  6=  -: •  (2) 

BC  sin  a 

In  the  triangle  DAC" , 

^      DA  ^      ^     tan  c  ,  ^ 

tanC=3^,ortanC=^j^.  (3) 

Combining  formulas  (2)  and  (3)  with  (i), 

^     tan  b  ,  . 

cos  C- .  (4) 

tan  a  ^  ' 

Again,  if  AB  were  made  the  base  of  the  right  spherical 
triangle  ABC,  we  should  have 

sin^=-: (5) 

s\x\a 

„     tan/^  •  (f.. 

tan  B—  — (o) 

sin  c 

r>     tan  c  f  . 

cosB= (7) 

tana  ^^ 

From   the   foregoing  equations  we  may  also  obtain   by 

combinations, 

cos^=sin  C  cos^.  (8) 

cosC— sin^  cosr.  (9) 

cos  «  =  cot  j5  cot  C  (10) 


NAPIER  S  RULES  OF  CIRCULAR  PARTS 
S3,  The   above   ten    formulas  are  sufficient  to   solve  all 
cases  of  right  spherical  triangles.     They  may,  however,  be 


RIGHT  AND   QUADRANTAL    TRIANGLES  95 

expressed  as  two  simple  rules,  called,  after  their  inventor, 
Napier's  rules. 

The  two  sides  adjacent  to  the  right  angle,  the  complement 
of  the  hypotenuse,  and  the  complements  of  the  oblique  an- 
gles are  called  the  circular  parts. 

The  right  angle  is  not  one  of  the  circular  parts. 


comp  a 


comp  0 


Thus  there  are  Jive  circular  parts — namely,  /',  c,  comprt,  comp 5,  compC. 
Any  one  of  the  five  parts  may  be  called  the  middle  part,  then  the  two  parts  next 
to  it  are  called  adjacent  parts,  and  the  remaining  two  parts  are  called  the  oppo- 
site parts. 

Thus  if  c  is  taken  for  the  middle  part,  comp^  and  b  are  adjacent  parts,  and 
comp  a  and  comp  Care  opposite  parts. 

The  ten  formulas  may  be  written  and  grouped  as  follows  : 


\st  Group. 
sin  comp  C=:tan  comp  a  tan  b. 
sin  comp  .5=tan  comp  a  tan  c. 
sin  comp  a  =tan  comp  j9  tan  comp  C. 
sin  <:  =rtan  compi?  tan  (5. 

sin  b  =tan  comp  C  tan<r. 


id  Group, 
sin  comp  a = cos  ^  cos  r. 
sin  (5=cos  coniprt  cos  comp^. 

sin  (-^cos  comp  a  cos  comp  C. 

sin  comp  i9  =  cos  comp  C  cos  b. 
sin  comp  C  =cos  comp  B  cos  c. 


Napier's  rules  may  be  stated  : 

I.  The  sine  of  the  middle  part  is  equal  to  the  product  of 
the  tangents  of  the  adjacent  parts. 

II.  The  sine  of  the  middle  part  is  equal  to  the  product  of 
the  cosines  of  the  opposite  parts. 


96 


SPHERICAL   TRIGONOMETRY 


84.  In  the  right  spherical  triangles  considered  in  this  work,  each 
side  is  taken  less  than  a  semicircumference,  and  each  angle  less  than 
two  right  angles. 

In  the  solution  of  the  triangles,  it  is  to  be  observed, 

(i.)  If  the  two  sides  about  the  right  angle  are  both  less  or  both 
greater  than  90°,  the  hypotenuse  is  less  than  90°;  if  one  side  is  less 
and  the  other  greater  than  90°,  the  hypotenuse  is  greater  than  90°. 

(2.)  An  angle  and  the  side  opposite  are  either  both  less  or  both 
greater  than  90°. 

EXAMPLE 


85.  Given  3  =  63°  56',  11^  =  40°  o',  to  find  c,  B,  and  C. 


To  find  c. 
comp  a  is  the  middle  part. 
c  and  h  are  the  opposite  parts, 
sin  comp«=cos(^  cosr, 
cos  a = cos  b  cosf. 
cos  a 

cos  C=- r- 

COS  0 

log  COS  a=9. 64288 
colog  COS  ^=0.11575 
log  cosr=g.75863 
^=54°  59'  47" 


To  find  C. 

comp  C  is  the  middle  part. 

comp  a,  and  b  are  adjacent  parts. 

sin  comp  C=tan  comprt  tan  b, 

cos  C=cot  a  tan  b. 


log  cot  rt=9.68946 
log  tan  3=9  92381 
9.61327 
C=65°  45'  58" 


To  find  B. 

b  is  the  middle  part. 

comp  a  and  comp  B  are  the  opposite 

parts. 

sin  (^=cos  comp  a  cos  comp  B 

or  sin  <5  =  sin  (z  sin  .5. 

■     u     ^^"  '^ 
sin  B  —  — — • 

sin  a 
log  sin  (^=9.80807 
colog  sin  c? =0.04659 
log  sin  ^5  =  9. 85466 

v5  =  45°4i'28" 

Check. 

Use  the  three  parts  originally  required. 

comp  C  is  the  middle  part. 

comp  .5  and  c  are  o]iposite  parts. 

sin  comp  C=cosr  cos  comp  B 

or    ■  cos  C=:cos  c  sin  B. 

log  cos ^=9. 75863 
log  sin  /^=9. 85466 
log  cos  C=g.  61329 

C=65°  45'  54" 


RIGHT  AND   QUADRANTAL    TRIANGLES 


97 


AMBIGUOUS  CASE 

56.  When  a  side  about  the  right  angle  and  the  angle  opposite 
this  side  are  given,  there  are  two  solutions,  as  illustrated  by  the  fol- 
lowing figure.  Since  the  solution  gives  the  values  of  each  part  in 
terms  of  the  sine,  the  results  are  not  only  the  values  of  a,  b,  B,  but 
i8o°— rt.  \ZcP—b,  iSo°—B. 


Given  c  =  26°  4'. 
C=36°o'. 


or 
or 


To  find  a,  a',  b,  b'  and  B,  B',  usin 
To  find  B  and  B' . 

sin  comp  C=  cos  comp  B  cosr, 
cos  C=sin  B  cos  c, 

.     -,     cos  C 
sinB=' 

cos  c 

log  cos  C=  9. 90796 

colog  cos  ^=0  04659 

log  sin  ^=9.95455 

B=  64°  14'  30" 
^'=i8o°-i5=ii5°  45'  30" 

To  find  b  and  />' . 
sin  3= tan  c  tan  comp  C, 
sin  ^=tan  c  cot  C. 
log  tan  ^=9. 68946 
log  cot  C=o.  13874 

log  sin  (5=:9. 82820 

(6=  42°  19'  17" 
(J'  =  l8o°  — i!'=i37°  40'  43" 


ig  Napier's  rules. 

To  find  a  and  a'. 

sin  r=cos  comp  a  cos  Comp  C, 
or         sin  ^=sin  a  sin  C, 

sin  c 

or         s\na=:— . 

sin  C 

log  sin  f  =  9.64288 

colog  sin  (7=0.23078 

log  sin  a =9. 87366 

a=  48°  22'  55"- 
a'  =  l8o°— 0=13103'/  5"  + 
(Discrepancy  due  to  omitted  decimals.) 

Check. 
sin  ^=cos  comp  a  cos  comp  B, 
sin  ^=sina  sin  B. 
log  sin  a  or  ^'  =  9. 87366 
logsini9ori5'  =  9.95455 
log  sin3=9. 82821 

b  =  42°  19'  21" 
<5'=  180°— ^=137040'  39" 


98  SPHERICAL    TRIGONOMETRY 

QUADRANTAL   TRIANGLES 
57.  Def. — A  quadrantal  triangle  is  a  spherical  triangle 
one  side  of  which  is  a  quadrant. 

A  quadrantal  triangle  may  be  solved  by  Napier's  rules  for 
right  spherical  triangles  as  follows : 

By  making  use  of  the  polar  triangle  where 

^  =  i8o°— a'  a  =  i8o°  — /4' 

^=180°—^'  ^=180°  — 5' 

C=i8o°  — c'  ^=180°— C 

we  see  that  the  polar  triangle  of  the  quadrantal  triangle  is 
a  right  triangle  which  can  be  solved  by  Napier's  rules. 
Whence  we  may  at  once  derive  the  required  parts  of  the 
quadrantal  triangle. 

EXAMPLE 

Given       A  =  136°  4'.  B  =  140°  o'.  a  =  90°  o'. 

The  corresponding  parts  of  the  polar  triangle  are 

a'  =  63°  56',  d'  =  40°  o'.  A'  =  900 

By  Napier's  rules  we  find 
^'  =  450  41' 28",  C'  =  65°45' 58",  ^=54°  59' 47"; 

whence,  by  applying  to  these  parts  the  rule  of  polar  triangles,  we 
obtain 

/5=  1 340  18' 32",  ^=114°  14' 2",  C=i25°o'i3". 

EXERCISES 

88.  (I.)  In  the  right-angled  spherical  triangle  ABC,  the  side  a= 
63°  56',  and  the  side  ^  =  40°.  Required  the  other  side,  c,  and  the 
angles  B  and  C. 

(2.)  In  a  right-angled  triangle  ABC,  the  hypotenuse  a  =  91°  42',  and 
the  angle  B  =  g^°  6'.     Required  the  remaining  parts. 

(3.)  In  the  right-angled  triangle  ABC,  the  side  b^^26°  ^',  and  the 
angle  B  =  ;^6°.     Required  the  remaining  parts. 

(4.)  In  the  right-angled  spherical  triangle  ABC,  the  side  ^=54°  30', 
and  the  angle  i?  =  44°  50'.     Required  the  remaining  [)arts. 

Why  is  not  the  result  ambiguous  in  this  case.' 


RIGHT  AND   QUADRANTAL   TRIANGLES  99 

(5.)  In  the  right-angled  spherical  triangle  ABC,  the  side  ^=55°  28', 
and  the  side  c=6t,°  15'.     Required  the  remaining  parts. 

(6.)  In  the  right-angled  spherical  triangle  ABC,  the  angle  B= 
69°  20',  and  the  angle  ^=58°  16'.     Required  the  remaining  parts. 

(7.)  In  the  spherical  triangle  ABC,  the  side  a  =  90°,  the  angle  C= 
42°  10',  and  the  angle  A  =  115°  20'.  Required  the  remaining  parts. 
Hint. — The  angle  A  of  the  polar  triangle  is  a  right  angle. 

(8.)  In  the  spherical  triangle  ABC,  the  side  b  =  90°,  the  angle  C= 
69°  13' 46",  and  the  angle  ^  =  72°  12' 4".  Required  the  remaining 
parts. 

(9.)  In  the  right-angled  spherical  triangle  ABC,  the  angle  C=  23° 
27'  42",  and  the  side  b  =  10°  39'  40".  Required  the  angle  B  and  the 
sides  a  and  c. 

(10.)  In  the  right  spherical  triangle ^^C,  the  angled  =  47°54'  20", 
and  the  angle  C=6i°  50'  29".     Required  the  sides. 


CHAPTER   IX 
OBLIQUE-ANGLED   TRIANGLES 

89,  Let  O  be  the  centre  of  a  sphere  of  unit  radius,  and 
ABC  an  obHque-angled  spherical  triangle  formed  by  the 
three  planes  A  OB,  BOC,  and  AOC.     Suppose   the   plane 


AED  passed  through  the  point  A  perpendicular  to  AO,  in- 
tersecting the  planes  AOB,  BOC,  and  AOC,  in  AE,  ED, 
and  AD  respectively.  Then  AD  — tan  b,  AE—\.ax\  c,  0D  = 
sec  b,  OE  —  ?,&cc. 

In  the  triangle  ROD, 

/t /9'^  =  sec V; -f- secV  —  2  stcb  stcc  cosa. 
In  the  triangle  AED, 

ED'—  tanVv' -|- tanV —  2  la.nb  tanr  cos  A. 
Subtracting  these  two  equations  and  remembering  that 
sec'^i^  —  tan°i5=i,  we  have 
0  =  2  —  2  secb  se.cc  cosa  +  2  tan^  tan^-  cos  A. 
Reducing,  we  have 

cosa=iC086  cosf  +  sin6  sine  cos^.  (i) 


OBLIQUE-ANGLED    TRIANGLES  lOf 

If  we  make  b  and  c  in  turn  the  base  of  the  triangle,  we  obtain  in  a 
similar  way, 

cos (^  =  cose  cos rt  + sine  sin  a  cos^, 
and  cose  =  cosa  cos(^  +  sina  sin^  cosC 

Remark. — In  this  group  of  formulas  the  second  may  be  obtained 
from  the  first,  and  the  third  from  the  second,  by  advancing  one  letter 
in  the  cycle  as  shown  in  the  figure ;  thus,  writing  b  for 
a,  c  for  b,  a  for  c,  B  for  A,  C  for  B,  and  A  for  C.  The 
same  principle  will  apply  in  all  the  formulas  of  Oblique- 
Angled  Spherical  Triangles,  and  only  the  first  one  of 
each  group  will  be  given  in  the  text. 

90,  By  making  use  of  the  polar  triangle  where 

d=iSo°-B'  B=iZo°-b' 

c=\?>(f-C  (;=i8o°-<;' 

we  may  obtain  a  second  group  of  formulas. 

Substituting  these  values  of  a,  b,  c,  and  A  in  (i),  and  remembering 
that  cos(i8o''— ^)  =  — cos/i  and  sin  (i8o°— /?)  =  sin^,  we  have 
cosA'  =  —  cosB'  cosC'  +  sini5'  sin  C  cos  a'. 
Since  this  is  true  for  any  triangle,  we  may  omit  the  accents  and 
write, 

cos  A=  -  cos  B  cos  C  +  sin  B  sin  C  cos  a.  (2) 

FORMULAS   FOR   LOGARITHMIC   COMPUTATION 
91»  Formula  (i),  cos«  =  cos^  cost  +  sin  ^  sine  cos/i, 
.      COSrt  — cos<^  COS^ 

gives  cos^=: ■. — 7 — ^- . 

°  sin  0  sin  c 

By§36,  cosA  =  i—2  s'ur' ^A 

,,^,  .    a,     ,        COS«  — cos  <^   cose 

Whence      i— 2sint^= -. — r— ^ , 

sin  o  sin  c  ^ 

.  51     .     cos/^cose+sin*^  sin  e  — cos<2 

or  sin  iA= ; ■, 

2  sin  a  sine 


0  SPHERICAL    TRIGONOMETRY 

_cos(^— r)— cosa 
*~      2  sin  b  sin^      ' 

.    a-Vb—c    .    a—b-\-c 
sin sin — 

2        2 

~  sin  b  sin  c  \3°) 

Putting 

a-\-b-{-c  ,       a-\-b—c  ,  a—b-\-c  , 

=s,  then z=.s—c,  and =:s—b, 

2  2  2  ' 


(I) 


,                     .    ,   ^        /sin  (s—b)  sin  (j— c) 
we  have  sin*^  =:\/ ^ — j—. — ^ ^• 

V  sin  ^  sin^ 

Since,  also,     cos^  =2cos'^y4  —  i, 
we  have,  similarly, 

^       Sin  b  sin  c 

Hence  tanj^^.  A!"  (^-»)  ^*"  (^-^). 

V  8in8  8in(«  — a) 

By  a  like  process,  formula  (2)  reduces  to 

,         ,/    —coHSeofi(S—A)  ,.-. 

tan*a=\/  — -.  (II) 

92.  If,  in  formula  I,  we  advance  one  letter,  we  have 

^^  /sin^^-£)_sii7(7p_ 

^       Sin  J-  sin  {s  —  b) 
And  dividing  tan^^  by  tan  ^5,  and  reducing,  we  obtain 
idin^A      s\n{s-'b) 
tan  ^ B~  sin  {s—a)' 

By  composition  and  division, 

ta.n  ^  A -\- ta.n -^  B      sin  (^  — <^)4-sin  (j— «) 
tan -I  y^— tan  ^  B~  sin(j— ^)— sin  {s—a)' 

By  §§  30,  38,  this  becomes 

fi\i\^{A-^B)  tan^c 


8in^(4— jB)      taii^(a  — fe) 


(III) 


OBLIQUE-ANGLED    TRIANGLES  T03 

Multiplying  tan^^  by  tan^^,  and  reducing,  we  obtain 

tan-|^  tan-|^      sin(j— g) 
I  ~"      sinj     " 

By  division   and   composition,  and  by  §§  30,  38,  this  be- 
comes 

co8-^(^4--B)  _      tan^c  .^^ 

co8-J(^  — £)~  tan^(a  +  6)*  ^      ' 

Proceeding  in  a  similar  way  with  formula  II,  we  obtain 


8in^(a  — 6)      tan^(^  — £)* 


(V) 


And  co8|(a  +  &)  cotjC  ,^^ 

co8i(a— 6)~tan^(^  +  B)*  ^      ' 

03 »  In  the  spherical  triangle  ABC,  suppose  CD  drawn  per- 
pendicularly to  AB,  then,  by  the  formulas  for  right  spher- 
ical triangles. 


In  triangle  ACD,  sin  p  =  smb  sin  A. 

In  triangle  BCD,  sin  /  =  sin  ^  sin  B. 

Whence  sin  a  sin  B  =  sin  b  sin  Ay 

sin  a      sin  6  /tttt\ 

or  = (VII) 

sin  A      sin  B 

Remark.— li  (A  +  B)>i8o°,  then  (a  +  ^)>i8o°,  and  if  (A-t-B)<i8o°, 
then  (a-H^)<i8o''. 


104 


SPHERICAL    TRIGONOMETRY 


94,  All  cases  of  oblique-angled  triangles  may  be  solved 
by  applying  one  or  more  of  the  formulas  I,  II,  III,  IV,  V, 
VI,  VII,  as  shown  in  the  following  cases. 

CASES 

(i.)  Given  three  sides,  to  find  the  angles. 

Apply  formula  I.     Check:  apply  V  or  VI. 

(2.)  Given  three  angles,  to  find  the  sides. 

Apply  formula  II.     Check  :  apply  III  or  IV, 

(3.)  Given  two  sides  and  the  included  angle. 

Apply  V  and  VI,  and  VII.    Check  :  apply  III  or  IV, 

(4.)  Given  two  angles  and  included  side. 

Apply  III  and  I V,  and  VII.     Check  :  apply  V  or  VI. 

(5,)  Given  two  angles  and  an  opposite  side. 
Apply  VII,  V,  and  III.     Check:  apply  IV, 

(6.)  Given  two  sides  and  an  opposite  angle. 
Apply  VII,  V,  and  I V.    Check  :  apply  III. 


EXAMPLE— CASE  (l) 
95*  Given  a  =  81°  10'  bzs6cP2o'  <:  =  1 12®  35' 

To  find  A,  B,  and  C. 


a=  81°  10' 
b  =  60P  20' 
(T  =112°  25' 

2J=253°55' 

S  =  12(P  57'  30" 

J— 0=45°  47'  30" 
s—6=6()°  37'  30" 
j--^  =  i4°  32'  30" 
log  sin  J.=9. 90259 
1<^  sin (^—«)=9. 85540 
log  sin  (j-  —  <^)  =  9. 96281 
log  sin(j—^)=:9. 39982 


To  find  A. 


V  sin  J  sin  (j— a) 
log  sin(^  — ^)=9.9628l 
log  sin  (j— f)=g.39982 

colog  sin  ^=0.14460 

colog  sin(j— a)=o.0974l 

2)19.60464 

log  tan ^^=      9.80232 

i.(4=32°23'  ig" 

.4=64046' 38" 


OBLIQUE-ANGLED    TRIANGLES 


105 


To  find  B. 

UniB=./^-''}f^'-'y 
V       sinjsm(j— ^) 

log  sin(j— a)=9.85540 

log  sin(j— f)=g.39g82 

colog  sin  ^=0.09741 

colog  sin  (j— <^)  =0.03719 

2)19.38982 

\ogtaa^B=      9.69491 

iB=26°2i'    6" 

^=52°  42'  12" 


To  find  C. 

tan  i  6"=  .    hm{s-a)sm{s-b) 

V        sin  5  sin(j— f) 

log  sin (j-<2)=9. 85540 

log  sin  (j— (J) =9. 9628 1 

colog  sin  j=o.  09741 

colog  sin  (j — <•) = o.  600 1 8 

2)20.51580 

logtan^C=      10.25790 

iC=  61°   5' 32" 

C=I220ll'    4" 


Check. 

Formula  V.  cot^  C=  '^-^(^rf)  '^^(-+^), 

/4=64°46'  38" 
^=52042'  12" 

A 


-B=12° 

^{A-B)^  6° 

<i=8i°  10' 

i  =6cP  20' 
a +^=141°  30' ;  ^(a+^)=70°  45' 
a— 6=  20°  50';  i(«— ^)=ioP  25' 


4'  26" 

2'  13" 

log  tan^f^— .ff)=9.02430 

log  sin  ^((7 +  3)  =  9  97501 

colog  sin^(rt  —  i^)=o. 74279 

cot^  (7=9.74210 

^  c=  61°  5'  32" 

C=I22<=  II'    4" 


EX\MPLE— CASE  (3) 
»6.  Given  a  =  78°  1 5'  <J  =  56°  20' 

To  find  A,  B,  and  c. 

\{a-Vb')-=tf  17'  30" 
|(a-^)=io='  57'  30" 
^C=6o° 

Formula  VI  may  be  written 

taniM+^)=^2ii(l!Z^!!IgjK, 

log  cos \(a  —  F)^=  9. 99201 

log  cot^C=  9-76144 

colog  cos ^(^  +  <^)=  0.41337 


log  tan \{A-\- E)=\o.  16682 

^(^+^)  =  55°44'36"- 
k{A-B)=  6°  47'    4" 
/f  =62°  31'  40" 
^=48°  57'  32"- 


C=I20P 


log  sin  ^(rt+^)=9. 96498 
log  cos  ^(rt'  +  (^) =9.58663 
log  sin^(rt  — 1^)=9.27897 
log  cos  ^  (a  — ^)= 9. 99201 
log  cot  ^(7=9.76144 


Tofind\{A~B). 
Formula  V  may  be  written 

tanM^-^)^'"^^.^\"A'?!*'^- 

logsin^((j-/0  =  9.27897 

log  cot  ^  (7=9.76144 

colog  sin^((Z  +/')  =  0.03502 

K^-^)=6°  57' 4" 


zo6 


SPHERICAL    TRIGONOMETRY 


To  find  c. 
From  Formula  VII,  sinf= 


sin  b  sin  C 


sin  B 


log  sin  b  =9.92027 
log  sin  C=9. 93753  , 
colog  sin  B-=o.  12249 

log  sin  f =9. 98029 
<-=I07°8' 


tan^r: 


Check. 
Formula  III  may  be  written 

%m\{A-\-B)  tan^(a— ^) 

%\xi^{A-B) 

log  sin  ^(^+7?)=  9.91725 

log  tan ^ (a  —  (ii)  =  g.28696 

colog  sin ^(^—i«')=  0.92762 

log  tanJf=io.  13183 

\c=  53°  33'  56"- 
r=io7°    i  51"  — 
(Dis:.epancy  due  to  omitted  decimals.) 


AMBIGUOUS  CASES 
97.  (I,)  Two  sides  and  an  angle  opposite  one  of  them  are  the 
given  parts. 

If  the  side  opposite  the  given  angle  differs  frojn  QcP  more  than  the 
other  given  side,  the  given  angle  and  the  side  opposite  being  either  both 
•  less  or  both  greater  than  (jicP,  there  are  two  solutions. 


(2.)  Two  angles  and  a  side  opposite  one  of  them  are  the  given  parts. 

If  the  angle  opposite  the  given  side  differs  from  90°  more  i/ian  the 
other  given  angle,  the  given  side  and  the  angle  opposite  being  either 
both  less  or  both  greater  than  90°,  there  are  two  solutions. 

Remark. — There  is  no  solution  if,  in  either  of  the  formulas, 


sin^= 


sin  A  sin  h 


sin  /'  sin  A 


sin  a       '  'A'wB 

the  numerator  of  the  fraction  is  greater  than  the  denominator. 


OBLIQUE-ANGLED   TRIANGLES 


107 


EXAMPLE— CASE  (6) 
95.  Given  a=40°  16'  ^  =  47°  44'  ^=52°  30' 

To  find  B,  B\  C,  C,  and  c,  c\ 


tan  5  c  ■■ 


To  find  B  and  B'. 

Formula  VII  may  be  written 

.    D     sin  ^  sin  b 

sm  B  = 

sma 

log  sin  yl  =9.89947 

log  sin  6  =  9.86924 

colog  sin  a  =  0.18953 

log  sin 5  =9.95824 

B=    65°  16' 30" 
5' =  114°  43' 30" 
To  find  c. 
Formula  IV  may  be  written 

cos\{A+B)  tan|(a  +  6) 

coihiA-B) 

logcosH^+-S)  =9.71326 

log  tan  I  (a +6)  =  9.98484 

colog  cos  \{A  —B)=  0.00270 

log  tan  5  c  =  9.70080 

ic=26°39'42" 
c  =  53° 19' 24" 
To  find  c' . 

log  cos  \  {A  +5  0=9  -0463 1 

log  tan  I  (a + ft)  =9.98484 

colog  cos^(^  —B')  =  0.06745 

log  tan  \c'  =  9.09860 

^c'=    7°   9'   9" 
c'=i4°i8'i8" 


To  find  C. 
Formula  V  may  be  written 

cot  ^  C  =  ^'"  2  <^°+^)  ^^"  K^  -B) 
sin  5  (a  — J) 

log  sin  §(«+*)  =    9-84177 
log  tan  \{A  —B)  =   9.04901  n  * 
cologsinJ(a  — 6)  =    1. 18633  n 
logcot^C=  10.07711 

|C  =  39°56'24" 
C=  79°  52' 48" 

To  find  C. 
logsinKa+&)  =    9-84177 
logtanK^-S')  =    9-78153  n 
colog  sin  |(a  —  6)  =    1. 18633  ^ 

log  cot  I C  =  10.80963 

hC'=    8°  48' 41" 
C'  =  i7°37'"" 

Check. 
Formula  III  may  be  written 


sin  ft  = 


sin  B  sin  c 


sinC 

log  sin  B  =  9.95824 

log  sin  f  =  9.90418 

colog  sin  C  =  0.00682 

log  sin  6  =  9.86924 

ft  =  47°  44' 


EXERCISES 
99,  (i.)  In  the  spherical  triangle  ABC,  the  side  a  =  124°  53',  the 
side  <5  =  3i°  19',  and  the  angle  ^  =  16°  26'.     Find  the  other  parts. 

(2.)  In  the  oblique-angled  spherical  triangle  ABC,  angle  A=  128° 
45',  angle  C=  30°  35',  and  the  angle  j5  =  68°  50'.    Find  the  other  parts. 
*Tlie  letter  "  n  "  indicates  that  these  quantities  are  negative. 


tt*  SPHERICAL    TRIGONOMETRY 

(3.)  In  the  spherical  triangle  ABC,  the  side  c  =  78°  15',  ^=56°  10', 
and  A  =  1 20°.     Required  the  other  parts. 

(4.)  In  the  spherical  triangle  ABC,  the  angle  ^  =  1250  20',  the  an- 
gle C=48°  30',  and  the  side  ^  =  83°  13'.  Required  the  remaining 
parts. 

(5.)  In  the  spherical  triangle  ABC,  the  side  c=^6fP  35',  b  =  39°  10', 
and  a  =  71°  15'.     Required  the  angles. 

(6.)  In  the  spherical  triangle ^5C.  the  angle  .<4  =  109°  55',  B=n6P 
38',  and  C=i2o°  43'.    Required  the  sides. 

(7.)  In  the  spherical  triangle  ABC,  the  angle  .<4  =  130"  5'  22",  the 
angle  0=^6°  45'  28",  and  the  side  ^  =  44°  13'  45".  Required  the  re- 
maining parts. 

(8.)  In  the  spherical  triangle  ABC,  the  angle  .,4  =  33®  15'  7",  B=s 
31°  34'  38",  and  C=  161°  25'  17".     Required  the  sides. 

(9.)  In  the  spherical  triangle  ABC,  the  side  c=ii2°  22'  58",  ^  = 
52°  39'  4",  and  a  =  89°  16'  53".     Required  the  angles. 

(10.)  In  the  spherical  triangle  ABC,  the  side  ^  =  76°  35'  36',  fi  = 
50°  10'  30",  and  the  angle  .,4  =  34°  15'  3".  Required  the  remaining 
parts. 

AREA  OF  THE  SPHERICAL  TRIANGLE 

100,  It  is  proved  in  geometry  that  the  area  of  a  spherical 
triangle  is  equal  to  its  spherical  excess,  that  is, 
area.^={A  -{-  B -\-  C — 2  rt.  angles)  X  area  of  the  tri-rectangular  triangle, 
where  A,  B,  and  C  are  the  angles  of  the  spherical  triangle. 
Hence 

area  _^  +  ^-f  C—iSo** 

surface  of  sphere  ~"  720** 

The  surface  of  the  sphere  is  47r/?*,  therefore 


area 


=  ^^( 1805 ) 


The  following  formula,  called  Lhuilier's  theorem,  simpli- 
fies the  derivation   of  {A-^  B+C  —  lSo°)  where  the  three 


OBLIQUE-ANGLED   TRIANGLES  109 

sides  of  the  spherical  triangle  are  given ;  in  it  a,  b,  and  c 
denote  the  sides  of  the  triangle,  and  2s=a+b  +  c. 

tan  (  A+B+C--lSO°\     Vtan  i  s  tan  i  (s-a)  tan  4  {s-6)  tan  h  (s-c). 

EXERCISES 

(i.)  The  angles  of  a  spherical  triangle  are,  ^=63°,  ^=84°  21', 
(7=  79° ;  the  radius  of  the  sphere  is  10  in.  What  is  the  area  of  the 
triangle? 

(2.)  The  sides  of  a  spherical  triangle  are,  ^=6.47  in.,  ^^  =  8.39  in., 
^=9.43  in. ;  the  radius  of  the  sphere  is  25  in.  What  is  the  area  of 
the  triangle? 

(3.)  In  a  spherical  triangle,  ^=75°  16',  ^5=39°  20',  ^=26  in. ;  the 
radius  of  the  sphere  is  14  in.     Find  the  area  of  the  triangle. 

(4.)  In  a  spherical  triangle,  <? =44 1  miles, /^=  287  miles,  C=38° 21' ; 
the  radius  of  the  sphere  is  3960  miles.     Find  the  area  of  the  triangle. 


CHAPTER   X 

APPLICATIONS    TO    THE   CELESTIAL   AND   TERRES- 
TRIAL  SPHERES 

ASTRONOMICAL   PROBLEMS 

101,  An  observer  at  any  place  on  the  earth's  surface 
finds  himself  seemingly  at  the  centre  of  a  sphere,  one-half 
of  which  is  the  sky  above  him.  This  sphere  is  called  the 
celestial  sphere,  and  upon  its  surface  appear  all  the  heavenly 
bodies.  The  entire  sphere  seems  to  turn  completely  around 
once  in  23  hours  and  56  minutes,  as  on  an  axis.  The  im- 
aginary axis  is  the  axis  of  the  earth  indefinitely  produced. 
The  points  in  which  it  pierces  the  celestial  sphere  appear 
stationary,  and  are  called  the  north  and  south  poles  of  the 
heavens.  The  North  Star  (Polaris)  marks  very  nearly  (with- 
in 1°  16')  the  position  of  the  north  pole.  As  the  observer 
travels  towards  the  north  he  finds  that  the  north  pole  of  the 
heavens  appears  higher  and  higher  up  in  the  sky,  and  that 
its  height  above  the  horizon,  measured  in  degrees,  corre- 
sponds to  the  latitude  of  the  place  of  observation. 

The  fixed  stars  and  nebulae  preserve  the  same  relative 
positions  to  each  other.  The  sun,  moon,  planets,  and  com- 
ets change  their  positions  with  respect  to  the  fixed  stars 
continually,  the  sun  appearing  to  move  eastward  among 
the  stars  about  a  degree  a  day,  and  the  moon  about  thir- 
teen times  as  far. 


APPLICATIONS  III 

The  zenith  is  the  point  on  the  celestial  sphere  directly 
overhead. 

The  horizon  is  the  great  circle  everywhere  90°  from  the 
zenith. 

The  celestial  equator  is  the  great  circle  in  which  the 
plane  of  the  earth's  equator  if  extended  would  cut  the  celes- 
tial sphere. 

The  ecliptic  is  the  path  on  the  celestial  sphere  described 
by  the  sun  in  its  apparent  eastward  motion  among  the  stars. 
The  ecliptic  is  a  great  circle  inclined  to  the  plane  of  the 
equator  at  an  angle  of  approximately  23|° 

The  poles  of  the  equator  are  the  points  where  the  axis 
of  the  earth  if  produced  would  pierce  the  celestial  sphere, 
and  are  each  90°  from  the  equator. 

The  poles  of  the  ecliptic  are  each  90°  from  the  ecliptic. 

The  equinoxes  are  the  points  where  the  celestial  equator 
and  ecliptic  intersect;  that  which  the  sun  crosses  when  com- 
ing north  being  called  the  vernal  equinox,  and  that  which  it 
crosses  when  going  south  the  autumnal  equinox. 

The  declination  of  a  heavenly  body  is  its  distance,  meas- 
ured in  degrees,  north  or  south  of  the  celestial  equator. 

The  right  ascension  of  a  heavenly  body  is  the  distance, 
measured  in  degrees  eastward  on  the  celestial  equator,  from 
the  vernal  equinox  to  the  great  circle  passing  through  the 
poles  of  the  equator  and  this  body. 

The  celestial  latitude  of  a  heavenly  body  is  the  dis- 
tance from  the  ecliptic  measured  in  degrees  on  the  great 
circle  passing  through  the  pole  of  the  ecliptic  and  the 
body. 

The  celestial  longitude  of  a  heavenly  body  is  the  dis- 
tance, measured    in    degrees   eastward  on   the  ecliptic,   from 


112  SPHERICAL    TRIGONOMETRY 

the  vernal  equinox  to  the  great  circle  passing  through  the 
pole  of  the  ecliptic  and  the  body. 

EXERCISES 
(i.)  The  right  ascension  of  a  given  star  is  25°  35',  and  its  declina- 
tion is  +  (north)  63°  26'.     Assuming  the  angle  between  the  celestial 
equator  and  the  ecliptic  to  be  23°  27',  find  the  celestial  latitude  and 
celestial  longitude. 


In  this  figure  AB  is  the  celestial  equator,  A  C  the  ecliptic,  P  the  pole  of 

the  equator,  P'  the  pole  of  the  ecliptic.  S  is  the  position  of  the  star,  and 
the  lines  SB  and  SC  are  drawn  through  P  and  P  perpendicular  to  .^^and 
AC.  AB  is  the  right  ascension  and  BS  the  declination  of  the  star,  while 
AC  is  the  longitude  and  SC  the  latitude  of  the  star. 

In  the  spherical  triangle  P PS,  it  will  be  seen  that  P' S  is  the  comple- 
ment of  the  celestial  latitude,  PS  the  complement  of  the  declination,  and 
P PS  is  90°  plus  the  right  ascension.  It  is  to  be  noted  that  A  is  the  ver- 
nal equinox. 

(2.)  The  declination  of  the  sun  on  December  21st  is  — (south) 
23°  27'.  At  what  time  will  the  sun  rise  as  seen  from  a  place  whose 
latitude  is  41°  18'  north  ? 

The  arc  ZS  which  is  the  distance  from  the  zenith  to  the  centre  of  the  sun 
when  the  sun's  upper  rim  is  on  the  horizon  is  90^  50'.  The  50'  is  made  up 
of  the  sun's  semi-diameter  of  16',  plus  the  correction  for  refraction  of  34'. 


AP  PLICA  riONS 


"3 


(3.)  The  declination  of  the  sun  on  December  21st  is  —  (south) 
23°  27'.  At  what  time  would  the  sun  set  as  seen  from  a  place  in  lati- 
tude 50°  35'  north  ? 


In  these  figures  P  is  the  pole  of  the  equator,  Z  the  zenith,  EQ  the  celes- 
tial equator.  AS\^  the  declination  of  the  sun,  ZS—Qf:P  50',  /"^^go^  +  dec- 
lination,  7*^=  90°— latitude.  The  problem  is  to  find  the  angle  SPZ.  An 
angle  of  15°  at  the  pole  corresponds  to  i  hour  of  time. 

GEOGRAPHICAL    PROBLEMS 

102»  The  meridian  of  a  place  is  the  great  circle  passing 
through  the  place  and  the  poles  of  the  earth. 

The  latitude  of  a  place  is  the  arc  of  the  meridian  of  the 
place  extending  from  the  equator  to  the  place. 

Latitude  is  measured  north  and  south  of  the  equator  from  0°  to  goP. 

The  longitude  of  a  place  is  the  arc  of  the  equator  extend 
ing  from  the  zero  meridian  to  the  meridian  of  the  place. 
The  meridian  of  the  Greenwich  Observatory  is  usually  taken 
as  the  zero  meridian. 

Longitude  is  measured  east  or  west  from  0°  to  180°. 

The  longitude  of  a  place  is  also  the  angle  between  the  zero  meridian  and 
the  meridian  of  tlie  place. 


114  SPHERICAL    TRIGONOMETRY 

In  the  following  problems  one  minute  is  taken  equal  to  one  geo- 
gfraphical  mile. 

(I.)  Required  the  distance  in  geographical  miles  between  two 
places,  D  and  E,  on  the  earth's  surface.  The  longitude  of  D  is  60° 
rs'  E.,  and  the  latitude  20°  10'  N.  The  longitude  of  E  is  115°  20'  E., 
and  the  latitude  37°  20  N. 


In  this  figure  AC  represents  the  equator  of  the  earth,  P  the  north  pole 
andy^  the  intersection  of  the  meridian  of  Greenwich  with  tlie  equator.  Ph 
and  PC  represent  meridians  drawn  through  D  and  E  respectively.  Then 
AB  is  the  longitude  and  BD  the  latitude  of  Z> ;  AC  the  longitude  and  C£ 
the  latitude  of  £. 

(2.)  Required  the  distance  from  New  York,  latitude  40°  43'  N., 
longitude  74°  o'  W.,  to  San  Francisco,  latitude  37°  48'  N.,  longitude 
122°  28'  W.,  on  the  shortest  route. 

(3.)  Required  the  distance  from  Sandy  Hook,  latitude  40°  28'  N., 
longitude  74°  i'  W.,  to  Madeira,  in  latitude  32°  28'  N.,  longitude  16°  55, 
W.,  on  the  shortest  route. 

(4.)  Required  the  distance  from  San  Francisco,  latitude  37*^  48' 
N.,  longitude  122°  28'  W.,  to  Batavia  in  Java,  latitude  6°  9'  S.,  longi- 
tude 106°  53'  E.,  on  the  shortest  route. 

(5.)  Required  the  distance  from  San  Francisco,  latitude  37°  48' 
N.,  longitude  122°  28'  VV.,  to  Valparaiso,  latitude  33°  2'  S.,  longitude 
71"  41'  W.,  on  the  shortest  route. 


CHAPTER   XI 
GRAPHICAL   SOLUTION   OF   A   SPHERICAL  TRIANGLE 

103,  The  given  parts  of  a  spherical  triangle  may  be  laid 
off,  and  then  the  required  parts  may  be  measured,  by  making 
use  of  a  globe  fitted  to  a  hemispherical  cup. 

The  sides  of  the  spherical  triangle  are  arcs  of  great  circles, 
and  may  be  drawn  on  the  globe  with  a  pencil,  using  the 
rim  of  the  cup,  which  is  a  great  circle,  as  a  ruler.  The  rim 
of  the  cup  is  graduated  from  o°  to  i8o°  in  both  directions. 

The  angle  of  a  spherical  triangle  may  be  measured  on  a 
great  circle  drawn  on  the  sphere  at  a  distance  of  90°  from 
the  vertex  of  the  angle.* 

Case  I.  Given  the  sides  a,  b,  and  c  of  a  spherical  triangle, 
to  determine  the  ajigles  A,  B,  and  C. 

Place  the  globe  in  the  cup,  and  draw  upon  it  a  line  equal 
to  the  number  of  degrees  in  the  side  c,  using  the  rim  of  the 
cup  as  a  ruler.  Mark  the  extremities  of  this  line  A  and  B. 
With  A  and  B  as  centres,  and  b  and  a  respectively  as  radii, 
draw  with  the  dividers  two  arcs  intersecting  at  C  (Fig.  i). 
Then,  placing  the  globe  in  the  cup  so  that  the  points  A  and 
C  shall  rest  on  the  rim,  draw  the  line  AC=b,  and  in  the 
same  way  draw  BC=  a. 

To  measure  the  angle  A  place  the  arc  AB  in  coincidence 

*  Slated  globes,  three  inches  in  diameter,  made  of  papier-mache,  and  held  in 
metal  hemispherical  cups,  are  manufactured  for  the  use  of  students  of  spherical 
trigiinometry  at  a  small  cost. 


ii6 


SPHERICAL  TRIGONOMETRY 


with  the  rim  of  the  cup,  and  make  Ah  equal  to  90°.  Also 
make  AF  in  AC  produced  equal  to  90°.  Then  place  the 
globe  in  the  cup  so  that  E  and  F  shall  be  in  the  rim,  and 
note  the  measure  of  the  arc  EF.  This  is  the  measure  of  the 
angle  A.  In  the  same  way  the  angles  B  and  C  can  be  de- 
termined. 


Case  II.  Given  the  angles  A,  B,  and  C,  to  find  the  sides 
a,  b,  and  c. 

Subtract  A,  B,  and  C  each  from  180°,  to  obtain  the  sides 
a',  b\  and  c'  of  the  polar  triangle.  Construct  this  polar  tri- 
angle according  to  the  method  employed  in  Case  I.  Mark 
its  vertices  A\  B',  and  C.  With  each  of  these  vertices  as 
a  centre,  and  a  radius  equal  to  90°,  describe  arcs  with  the  di- 
viders. The  points  of  intersection  of  these  arcs  will  be  the 
vertices  A,  B,  and  C  of  the  given  triangle.  The  sides  of 
this  triangle  a,  b,  and  c  can  then  be  measured  on  the  rim 
of  the  cup. 


GRAPHICAL  SOLUTION' 


117 


Case  III.  Given  two  sides,  b  and  c,  and  the  included  angle 
A,  to  find  B,  C,  and  a. 

Lay  off  (Fig.  3)  the  line  AB  equal  to  c,  an  J.  mark  the 
point  D  in  AB  produced,  so  that  AD  equals  90°.  With  the 
dividers  mark  another  point,  F,  at  a  distance  of  90°  from  A. 
Turn  the  globe  in  the  cup  till  D  and  F  are  both  in  the  rim, 
and  make  DE  equal  to  the  number  of  degrees  in  the  angle  A. 
With  A  and  E  in  the  rim  of  the  cup,  draw  the  line  AC  equal 
to  the  number  of  degrees  in  the  side  b.  Join  C  and  B.  The 
required  parts  of  the  triangle  can  then  be  measured. 


.Jk- 

^m. 

Ip 

w    ^m 

P:  ..  .ym^ 

\ 

G 

r-'"^ 

p 

1  ■          ,  :-.:-/■  .:-.-\-    , 

FIG.    3 


FIG.    4 


Case  IV.  Given  t/ic  angles  A  and  B  atul  the  included  side 
c,  to  find  a,  b,  and  C. 

Lay  off  the  line  AB  equal  to  c.  Then  construct  the  given 
angles  at  A  and  B,  as  in  Case  III.,  and  extend  their  sides  to 
intersect  at  C. 

Case  V.  Given  the  sides  b,  a,  arid  the  angle  A  opposite  one 
of  these  sides,  to  find  c,  B,  and  C.     (Ambiguous  case.) 


ii8  SPHERICAL    TRIGONOMETRY 

Lay  off  (Fig.  4)  AC  equal  to  b,  and  construct  the  angle  A 
as  in  Case  III.  Take  c  in  the  dividers  as  a  radius,  and  with 
6^  as  a  centre  describe  arcs  cutting  the  other  side  of  the  tri- 
angle in  B  and  B\  and  measure  the  remaining  parts  of  the 
two  triangles. 

If  the  arc  described  with  C  as  a  centre  does  not  cut  the  other  side  of  the 
triangle,  there  is  no  solution.     If  tangent,  there  is  one  solution. 

Case  VI.  Given  the  angles  A,  B,  and  the  side  a  opposite 
one  of  the  angles. 

Construct  the  polar  triangle  of  the  given  triangle  by 
Case  V. ;  then  construct  the  original  triangle  as  in  Case  II., 
and  measure  the  parts  required. 

The  constructions  given  above  include  all  cases  of  right  and  quadrantal 
triangles. 


CHAPTER   XII 
RECAPITULATION   OF   FORMULAS 

ELEMENTARY  RELATIONS  <'§  lO) 

cos  X 

cot  X  = , 

sin^ 

I 


tan^c 

:= 

sin;v 

cos:*: 

secx 



I 

cos  X  sin  X 

tan^  cot;c=  i, 
sin^jc  +  cos^x  =  I, 
I  +  tan^^  =  sec^x, 
T  +  cot^jc  =  csc^  j;. 

RIGHT  TRIANGLES  (§§14  AND   27) 

sin^  =  -,  sin^  =  -, 

c  c 

cosA  =  -,  cosB  =  ~,  ^ 

c  c 

tan^=-,  tan^  =  -, 

b  a 

h  /T 

COt^  =  -,  C0tj9  =  -, 

a  b 

where  c  =  hypotenuse,  a  and  b  sides  about  the  right  angle  ;  A  and  B 
the  acute  angles  opposite  a  and  b. 

FUNCTIONS  OF  TWO  ANGLES  (§§  3O-34) 

sin  (x  -\-y)  =  sin  x  cosy  +  cos  x  siny, 
sin(:x:— _y)  =  sin^  cosj  —  cos:f  sin_y, 
cos  {x  -\-}')  =  cos  X  cosy  —  sin  x  sin  j, 
cos  (jx  —  y)  —  cos  x  cosy  +  sin  x  sinjy. 


120  RECAPITULATION  OF  FORMULAS 

tan^r  +  tanj)/ 

tan  ix  +  y)  = ' '^— , 

^     '  -^        I— tan  ^  tan  V 

tan  (x—y)  — 
COt{x-\-y)  = 
cot  {x—y) 


tan^ 
tan;r  —  tan^ 
i  +  tan;r  tuny' 
cot^  cot^ —  I 

cotj/  +  cotjr  ' 
cot  jr  cot  J/ 4- 1 
cot/  —  cotjr 


FUNCTIONS   OF  TWICE  AN  ANGLE  (§  36) 


sin  2^=:  2  sin;tr  cos;r, 
cos  2x  =  cos''jr  —  sin'jr, 
=  1  — 2  sin";!", 
=  2cos';r — I, 
2  tan;i: 


tan  2x  = 


cot  2 jr  = 


I  — tan-'jir 
cot^-r — I 


2  cot;r 

FUNCTIONS  OF  HALF  AN  ANGLE  (§  37) 

.     1          .       /i — cos^ 
sin  ix=^±:K/ , 


1           ■       /l-}-cos;jr 
cosi^=±^-X_ 

1  ,      /i — cosx 

tan  ^x  =  ±\    — ; , 

V  I  +  cos  X 

,  / '  +  <^os  X 

cot  *^=\/ . 

^        V  I  —  cos  X 


SUMS   AND    DIFFERENCES   OF   FUNCTIONS   (§  38) 

sin  2^ -(- sin  7/ =  2  sin  |(«-j-?/)  cos  ^(«  —  v), 

sin  t(  —  sin  7/:=  2  cos^(«-|-7')  sin  \{u  —  v), 
cos  u  -\-  cos  7y  =  2  cos  \{u-\-v)  cos  |  {u  —  v), 
cos  i(  —  cos  7/=  —  2  sin  \{u-^v)  sin  i^(?^  —  z/). 

sin  ?^-|-sin  7/ tan  ^  (//  -|-  7^) 

sin  u  —  sin  v     tan  ^{u  —  v)' 


RECAPITULATION  OF  FORMULAS  i2l 

r 
tf 

i  OBLIQUE  TRIANGLES  (§§  42-45) 

a_sinA  a _s\nA  b     sin  ^ 

^""sin^'  £-~sinC'  ~c~  ^\nC' 

a~  b     X.2.n\{A  —  B) 
a-^b~~x.?iX\\{A-\-BY 
a  —  c     tan^(^  — C) 
a-\-c~\.?in\{A^Cy 
b  —  c_x.Sin\{B  —  C) 
\h-\-c~X.z.n\{B-irC)' 

^t=.c^-\-a^~ 2ca  cos B, 
c^  —  d'^b'^  —  zab  cos  C 

s{s  —  a) 
s{s  —  b) 


wnere  5= ■ 


2 

tani^=j^.  tani5=j^.  tanJC=J~. 

where  Ar=\/^3gE5fc£). 

AREA   OF  A   TRIANGLE  (§  46) 

S—\ac  sin  B.      S=\ba  sin  C.      S=^b  sin  /j.  ^  -^ 

5=  -\/:r(5  — «)(j  — ^)(5— f). 

LOGARITHMIC,  COSINE,   SINE,   AND   EXPONENTIAL  SERIES 

(§58) 

^og^  (I  +  ^)  =^  — 7  "*"  T  ~  " +' etc. 

JT^         X'^        X* 

cosr=i-^-h-,-^-H,etc. 


122  RECAFITULATION  OF  FORMULAS 

sin  X  —  X \- — -  +  ,  etc. 

3!      5!      71 

^=i+;c-|--^  +  '^  +  ^  +  ,etc. 
2!      3!      4! 

DE   MOIVRE'S  THEOREM   (§  71) 
(cos  X  +  V—  I  sin  jc)"=  cos  7ix  -+-  V—  i  sin  nx. 

«-i       •            n(n—  i)(n—  2)       n--i       ■   ■,      . 
smnx  =  /icos       jc  sin  jr ^^ ^ ^cos     "*  jc  sm^  jr +,  etc 

3! 

«         nin—  i)        «-2       •  9      , 
cos  nx  —  n  cos  x ^^ ^  cos       x  sin''  :f  + ,  etc. 

2  ! 

HYPERBOLIC   FUNCTIONS   (§75) 

.r  —X 

sinh  ;c  = , 

2 


cosh  ^  =  — , 


2 
e^^  cos  :v  -f  /sin  x. 

ix  -  ix 

e    —  e 

sin  x= , 

2  / 

e    +  e 
cos  .V  = — —  • 


sin  zj;  =  -^^ -^  =  /  Sinn  x, 

2 

X     ,  —X 

cos  IX  =  — =  cosh  X. 

2 

SPHERICAL  TRIANGLES 
RIGHT    AND    QUADRANTAL    TRIANGLES    (§§  8^,  87) 

Use  Napier's  rules. 

OKLlgUE    TRIANGLES    (§§  89-93) 

COS  a  =  cos  d  COS  c  -\-  sin  ^  sin  r  cos  A. 
cos  A  =  —  cos  B  COS  C+  sin  B  sin  6'  cos  a. 

,     ,         [sin  (i-  — /^)  sin  (i-  — (-) 


>       sin  y  sin  (s  —  a) 


tan 


RECAPITULATION  OF  FORMULAS  123 

J     _     /    —  cos  ^  cos  (.S"  — yj) 

tan  2  «  - \^^3  (^  _  ^)  ^Q3  (^  _  ^• 

sin  \{A  ■\-  B)  _       tan  ^  <: 
sin  \{A  —  B)~  tan  i  (a  —  ^) 
cos  \{A  ■\-  B)  __       tan  i  r 
cos  i  (^  -  ^)      tan  1  (a  +  <^)* 

sin  \  {a-\-b^  _       cot  \  C 

sin  \{a  —  b)  ~  tan  i  (^  -B) 

cos  1^  (<?  +  <^)  _       cot|-  C 
cos  1  (a  -'  /^)  ~  tan  \{A+B) 

sin  a  sin  b 

sin  ^      sin  B 

AREA    OF    SPHERICAL   TRIANGLES    (§   lOl) 

area  =  ttK^    — ■ — 

V        180°        J 

rA+B+C-i?,o°\  ^  Vtan|j-tani(j-«)tani(j-/?')tan^(j-^). 


APPENDIX 

RELATIONS  OF  THE  PLANE,  SPHERICAL,  AND  PSEUDO- 
SPHERICAL  TRIGONOMETRIES 

We  have  up  to  the  present  considered  the  trigonometries 
which  deal  with  figures  on  a  plane  or  spherical  surface.  A 
characteristic  feature  of  these  two  surfaces  is  that  the  curv- 
ature of  the  plane  is  zero,  while  that  of  the  sphere  is  a  posi- 
tive constant  p.  If  the  radius  of  the  sphere  is  increased  in- 
definitely, its  surface  approaches  the  plane  as  a  limit  while 
its  curvature  p  approaches  o. 

In  works  on  absolute  geometry  it  is  shown  that  there  ex- 
ists a  surface  which  has  a  constant  negative  curvature :  it  is 
called  a  pseudo-sphere,  and  the  trigonometry  upon  it  pseudo- 
spherical  trigonometry. 

We  observe  that  as  p  passes  continuously  from  positive 
to  negative  values,  we  pass  from  the  sphere  through  the 
plane  to  the  pseudo-sphere.  Thus  the  formulas  of  plane 
trigonometry  are  the  limiting  cases  of  those  of  either  of  the 
two  other  trigonometries. 

In  the  treatment  of  spherical  trigonometry  the  radius  of 
the  sphere  has  been  taken  as  unity.  If,  however,  the  radius 
of  the  sphere  is  r,  and  a,  b,  and  c  denote  the  lengths  of  the 
sides  of  the  spherical  triangle,  the  formulas  are  changed,  in 

that  a  is  replaced  by  -,   b  by  -,   and  ^  by  -  ;  thus, 


126  APPENDIX 

.    ^     sine 

smC=-: — 

sin  a 

.   e 
sin- 

becomes  sin  C= • 

.  a 
sm- 

r 

The  formulas  for  pseudo-spherical  trigonometry  are  the 
same  as  the  formulas  of  spherical  trigonometry,  except  that 

the  hyperbolic  functions  of  -,   -,  and  -  are  substituted  for 

the  trigonometric. 

Thus,  corresponding  to  the  above  formula  of  spherical 
trigonometry,  is  the  formula 

sinh- 

smC= 

sinh- 
r 

of  pseudo-spherical  trigonometry. 


PSBUOO-SPHBRE 


The  pseudo-sphere  is  generated  by  revolving  the  curve  whose  equation  is 


y=r  log -y/r*-,,  -i 

about  its  y  axis.     The  radius  of  the  base  of  the  pseudo-sphere  is  r. 


APPENDIX  127 

Hence  the  formulas  of  plane  trigonometry  can  be  derived 
from  the  formulas  of  either  spherical  or  pseudo-spherical 
trigonometry  by  expressing  the  functions  in  series  and  allow- 
ing r  to  increase  without  limit. 

Example. — Show  that  if  r  is  increased  indefinitely  the  following 

corresponding  formulas  for  the  spherical  and  pseudo-spherical  right 

triangle 

a  b        c  /  N 

cos  -  =  cos -cos -,  (i) 

r  r        r 

cosh  -  =  cosh  -  cosh  -  (2) 

r  r  r'' 

reduce  to  the  corresponding  formula  for  a  plane  right  triangle  ;  that 
is,  to 

^2  =  ^'  +  ^.  (3) 

Substituting  the  series  cos  -,  etc.,  in  equation  (i),  we  obtain 

(-^(:r--H-Mr— )(-f,(-:r— )• 

or  I  _  +  -I-  .  .  .  =  I  _  +  +  .  .  .  (4) 

2  !  r^      4  1  /"  2  !  ;-      2 !  r-'      4  !  r< 

Substituting  in  equation  (2)  the  series  for  cosh      ,  etc.,  which  we  obtain  from 

.X    \        —  X 

cosh  X  = —  ,  we  have 


2 

I  +• 


wr---hm--){'^m^-} 


or  1  -i 1 |-.--  =  i-| -\ 1 f----  (5) 

2I  r-      4 !  r*  2l  r^      2 !  r^      4 !  r* 

Cancelling  i  in  equations  (4)  and  (5),  multiplying  by  r^,  and,  finally,  allowing 
r  to  increase  without  limit,  we  get  from  either  equation 

a-  =6^  +  c\ 

EXERCISES 

Derive  each  of  the  following  formulas  of  plane  trigonometry  from 
the  corresponding  formula  of  spherical  trigonometry,  and  also  from 
the  corresponding  formula  of  pseudo-spherical  trigonometry. 


128  APPENDIX 

Right  triangles  ;  A  =  right  angle, 
(i.)  Plane  sinC=f 

Spherical,  sin  C  ■■ 


a 
sin  c 

sin<z 


Pseudo-spherical,      sin  C  =■  — —  • 

ainha 

ObUque  Triangles. 

(2.)  Plane,  a^  ^  IP- ■\-  <?■  —  2  be  cos  A. 

Spherical,  cos  a  =  cos  b  cos  ^  +  sin  ^  sin  e  cos  A. 

Pseudo-spherical,  cosh  a  =  cosh  b  cosh  e  +  sinh  b  sinh  <r  cos  A. 

(3.)  Plane,  6"=  Vj(j— a)(j'  — <^)(j— ^).  ^ 

Spherical, 

U„ (fl+5+C- 180:1  ^  Xniita„ife=i?lmni(£=^.ani(£lli). 
4  ^"/-""r  "r  r 

Pseudo-spherical, 


tanll8o:il^l±^±^  =  Xrihi^tanhi(£:^tanh^(£:i^tanhji£^, 
4  '         V  r  r  r 


ANSWERS   TO    EXERCISES 


§  4  (page  3). 

(I.)  i92°si'25f". 
Quadrant  III. 

(2.)    25°. 

(3.)    287°,  647°. 

(4.)  Quadrant  III. 

§  9  (page  9). 

tan  1000°  is  negative, 
cos  810°  is  o. 
sin  760°  is  positive, 
cot  —  70°  is  negative, 
cos  —  550°  is  negative, 
tan  —  560°  is  negative, 
sec  300°  is  positive, 
cot  1560°  is  negative, 
sin  130°  is  positive. 
cos  260°  is  negative, 
tan  310°  is  negative. 

§  13  (page  11), 
(3.)  cos  -30°=  2-V3, 
tan  -30°==- jV3> 
cot  -3o°  =  -V3, 
sec  -30°=  I  V3, 

CSC   —  ^0°  =  —  2. 

(4.)  cosjc  =  — IV2, 
tan  x=  \^2, 
cot  X  =  2  V2, 
sec  X  =  —  ^  V2, 
CSC  .r  =  —  3. 


(5.)  cos;'  =  |,     tan  ;/  =  -!, 

cotjK  =  — |,     sec;/  =  |, 

CSC  J  =  —  f . 
(6.)  sin6o°  =  i-V3, 

tan  60°  =  V3, 

cot  60°  =  ^  V3, 

sec  60°=  2, 

CSC  60°  =  I  V3. 
(7.)  coso°=i,     tano°=o. 
(8.)  sin  2  =  4,     cos  3  =  I, 

cot  2=1,     sec  2  =  f , 

CSC  z  =  f . 

(9.)  sin  45°  =  cos  45°  =  1 V2, 

tan  45°=  I, 

sec  45°=  CSC  45°=  V2. 
(10.)  sin^y^— ^V5,  cosji'=— f, 

cot_>;=|V5,     secj=-|, 

csc;'=-f  V5. 
(i  I .)  sin  30°=  i  cos  30°=  I V3.. 

tan3o°=  LV3, 

sec  30°=  I V3, 

CSC  30°=  2. 

(12.)  sinjc=4,     cos:r  =  — f. 
(13.)  Vf±iV5.. 


§  17  (page  14). 
(i.)  sin  70°=  cos  20°, 
cos  60°  =  sin  30°, 
cos  89°  31'  =  sin  29', 
cot  47°  =  tan  43°, 


13© 


ANSWERS    TO  EXERCISES 


tan  630=  cot  27° 

sin  72°  39'=  cos  17°  21'. 

(2.)  X  =  30°. 

(3.)  X  =  22°  30'. 

(4.)  .r=i8« 

(5.)  ^=150 

§  25  (page  21). 

(I.)  225°  and  3 1 50, 
60°  and  240°. 
(2.)  60°,  120°,  420P,  480P. 
(3.)  sin— 30°=— f 

cos— 3oo=i'/3. 
sin  765°=  cos  765  =  J  -/I, 
sin  120°=^  y'3, 
cos  1 20°  =  —  i, 
sin  210°= — J, 
cos  210°=  —  ^  -v/3. 
(4.)   The   functions  of  405°  are 
equal  to  the  functions  of  45°, 
sin  600°=  —  ^  'v/3, 
cos  600°=— I, 
tan  600°=  -y^3, 

cot  600°=  J  y/2, 
sec  600°=  —2, 
CSC  600°=  — f  V3. 

The  functions  of  1125°  are 
equal  to  the  functions  of  45°. 
sin  — 45°=— ^-/2, 
cos  — 45°=i -/2. 
tan  —  45°=  cot  —  450=  —  I , 
sec  —  45°=  -v/2, 
CSC— 45°=  — y'J. 
sin  225°=  cos  225°=—^  -/J, 
tan  325°=  cot  225°=  I, 
sec  225°=  CSC  225==  —  v/2. 
(5.)  The  functions  of  —  120°  are 


the  same  as  those  of  600*' 

given  in  (4). 

sin  ~  2250  =  I  -v/2, 

cos  —  225°  = — I  -v/2, 

tan  —  225°=  cot  —  225°=  —  I . 

sec  —  225°=  —  -v/2, 

esc  — 22  5°= -v/2, 

sin  — 420°  =  — ^  -/5, 

cos  —  420^  =  ^, 

tan  —  420°  =  —  y  J, 

cot  —  420°  =  —J  -v/3, 

sec  —  420°  =  2, 

CSC  —  420°  ==  —  f  -v/3! 

The  functions  of  3270°  are 

equal  to  the  functions  of  30°. 
(6.)  sin  233°  =  — cos  370 

cos  2330  =  — sin  37° 

tan  233°  =  cot  37° 

cot  233°  =  tan  370, 

sec  233°  =  —  CSC  27°, 

CSC  233°  =  —  sec  37". 

sin  — 197°  =  sin  17° 

cos  — 1970  =  — cos  17°, 

tan— 197°  =  — tan  17°, 

cot  —  1 970  =  —  cot  1 70 

sec— 197°  =  — sec  17°, 

CSC—  197°  =  CSC  170. 

sin  894°  =  sin  6° 

cos  894°  =  —  cos  6°, 

tan  894°  =  —  tan  6° 

cot  894°  =  —  cot  6°, 

sec  894°  =  —  sec  6°, 

CSC  894°  =  esc  6°. 
(7.)  sin  267°  =  — sin  870, 

tan  —2540=  —  tan  74°, 

cos  950''  =  —  cos  50°. 
(8.)  —0.28. 


ANSWERS   TO  EXERCISES 


131 


(9.)  2  sin*  X. 

(io.)sec'^:r  —  I. 

(II.)  sin  (;r  — 90°)=— cos^, 
cos  {x — 90°)  =  sin  X, 
tan  {x  —  90°)  =  —  cot  X, 
cot  {x — 90°)  =  —  tan  X, 
sec  (x  —  90°)  =  esc  x, 
CSC  (x — 90°)  =  —  sec  X. 

§  28  (page  24), 

(I.)  a  =62.324, 

A  =  32°  52'  40". 
(2.)    ^  =  21.874, 

^  =  390  45' 28", 

^=50°  14'  32". 
(3.)    a  =  300.95, 

/J  =  683.96, 

^  =  66°  15'. 
(4.)        ^  =  26.608, 
c  =  45763, 
^=35^33'. 

area  =  495-34. 

(5.)       ^  =  3-9973. 
^  =  4.1537, 

A  =  is° 46'  33". 
area  =  2.257. 
(6.)  ^  =  0.01729, 
(7.)  a  =  298.5.  , 
(8.)  yi  =  39°  42'  24". 
(9.)  r  =  2346.7. 
(ic.)  /y  =  28°  57' 8". 
(II.)  444.16  ft. 
(12.)  186.32  ft. 

(13.)  34°  33'  44". 
(14.)  303.99  ft. 

(1 5-)  238.33  ft. 

(16.)  15  miles  (about). 

(I7-)  79079  ft. 
(l8.)  165.68  ft. 


(I9-)  53°  33'- 
(20.)  II 5.136  ft. 
(21.)  76.355  ft. 
(22.)  ^  =  80°  32", 

^  =  C  =  49°59'44". 
(23.)      ^=53°  16' 36", 
i>=  12.0518  in., 

area  =  72.392  sq,  in. 
(24.)       <J=  130.52  in., 

area  =  24246  sq.  in. 
(25.)  23.263  ft. 
(26.)  17°  48". 
(27-)  5-3546  in. 
(28.)  1084950  sq.  ft. 
(29.)  17  ft.,  885  sq.  ft. 
(30.)       radius  =  24.882  in., 

apothem  =  2o.i3  in., 
area  =  1472  sq.  in. 
(31.)  12.861. 
(32.)  1782.3  sq.ft. 
(33.)  38168  ft. 
(34.)  20.21  ft. 
(35-)  2518.2  ft. 

§  29  (page  28). 

(I.)  ^  =  22°  58', 
^  =  7.07, 

c  =  9.0046. 
(2.)    <J  =  79-435. 
^=45°  27'  14", 
C  =  95°  24' 46". 
(3.)      ^i?  =  7.6745, 
^^'  =  2.6435, 
B  =  46°  43'  50", 
B'=  133°  16'  10", 
ACB  =  ios°  53'  10", 
ACB'  =  19°  20'  50". 
(4.)  A  =  37°  S3'. 
^  =  430  53 '25'', 


'32 


ANSWERS   TO  EXERCISES 


(5.)  902.94. 
(6.)  1253.2  ft. 

(7.)  357.224  ft: 

(8.)       ^=44°  2' 9", 
5  =  51°  28'  II", 
C  =  840  29'  40", 
area=  126 100  sq.  ft. 
(9.)  407.89  ft. 
(10.)  Z>'=i2i°  21'  16", 
C=92°6'38". 
D  =  7i°  II' 6". 

(II.)  5C=  5-672, 

DC=  3-694- 

§  34  (page  34). 

(2.)  sin  (450-1- .t_)  = 

\-\/2  (cos.r-|- sin  ^), 

cos  (45°+  ^')  = 

\  yj z  (cos.r  —  sin  jr), 
sin  (30°—  xj  = 

\  (cosjr—  -v/3  sin  jr), 

\  ("v/s  cos.r-}-sin;r), 
sin  (6o°-f--r)  = 

\  (\/3  cos.v  fsinx), 

cos  (6o°-f  -r)  = 

I  (cos  X  —  \/3  sin  -r). 
(3.)  sin(;r+>')  =  ff. 
sin(j'— j)=^. 

,    .        „     1/6-1-^/2 
(4.)  sin  75°  = 

cos  75°  = 
(5.)  sin  15°= 


cos  15°= 


_4 
-\/6  — \/2 

4 
-v/6  — 'v/2 
4  ■ 

\/6+  -/a" 


§  39  (page  37). 

(5.)  sin(45°-.r)  = 

i  -v/S  (cos  X  —  sin  JT), 

cos(45°— ^)  = 

\  yfi  (cos  ;r  -|-  sin  x), 

sin(45°-f^j  = 

I  'V/2  (cos  X  -\-  sin  jr), 

cos  (45°+-^)  = 

h  \/2  (cos  -tr — sin  x). 

(6.)  tan  7 5°  =2-1- -v/3. 
tan  15°  =  2  — -v/3. 


(14.)  sini/ 


cosij 


3+V5 


6      ' 


(15.)  sin  2X  =  —^, 
cos  2Jr  =  —^5. 


(16.)  sin  22i°=iy/2— -v/2, 
COS  22^°  =  I  iy/2  -H  v/2, 

tan  22^°=  v/2  — I, 
cot  22|°  =  -v/2-(- 1, 
sec  22^°  =  a/4  — 2-V/2, 

CSC  22^°  =  \/4 -|- 2 -v/2. 

-v/5  — I 

(17.)  X^. 

(18.)  sin  T5°  =  s\/^~-/3. 

cos  I  5°  =  J  >y2  -I-  -v/3, 


ANSWERS    TO  EXERCISES 


m 


tan  i5°  =  2  — -/i, 
cot  1 5°  =  2  +  Vs. 

sec  i5°=:2W2  — -y/s. 


CSC  i5°  =  2y^2  +  v/3- 

(20.)  sin  5Jr  = 

5  sin  X  —  20  sin^  x 

-|-i6  sin*x. 
(21.)  cos  $xr= 

5  cos  ;tr  —  20  cos'  X 

-j-i6  cos';r. 
(23.)  The  values  of  ;r  <  360°  are 
0°  30°,  1 50^  1 80°,  2 1 0°,  330°, 
(36.)  tan^"  tan/. 

§  41  (page  40). 

(I.)  sin-«  ^  -\/2  =  450,  135° 

45°+ 360°,  etc. 
cos— I  I  =60°  300°,  etc., 
tan  —  '  (—1)=  135°,  31 5°  etc. 
cos— »  I  =0°  360°,  etc., 
sin  —  I  (—  J)  =  2 10°,  330°,  etc 

(2.)  tan.r  =  3. 

(3.)  cos;r  =  ±|,  tanx  =  dz|. 

(4.)  sin(tan-'i-/3)=±i- 
(5.)  sin  (cos -I  I)  =  ±1. 

(6.)  cot(tan-'xV)=i7- 

(8.)  45°.  225°. 

(9.)  ;r  =  45°./=l8oo. 

(10.)  sin~'a  =  225°. 

§  48  (page  46). 

(I.)  C=i2i°33', 
^  =  2133.5, 

<r  =  2477.8. 

(2.)  C=55°4i'. 

^=534.05. 


^  =  653.52. 
(3.)  C=45°34'. 

«  =  i548.i, 

^=1293.7. 
(4.)  ^  =  105°  59', 

a=i  54.018, 

r=:  47.738. 

(5.)  ^  =  68°  58', 
^  =  5274.9, 

(6.)  i?=54"58'. 
«  =  923.4, 
c=  1 187.7. 


§  49  (page  47). 


(I.)  (I.) 
(2.) 

(3-) 

(4-) 

(2.)  i?  = 

C- 

C  — 

(3-)  c^ 
B- 
C  = 

(4.)<r  = 
B  = 
C  = 

(5.)  No 

(6.)  b  = 
A= 
B  = 


Two  solutions. 

One  solution,  a  right  tn 

angle. 
One  solution. 
Two  solutions. 
1 16°  57'  21", 
- 1 5°  50'  39"' 

:  0.32122. 

=  2.5719, 

=  13°  15'  l", 

=  142°  13'  59". 

93.59,  ^=12.07, 

26°  52' 7", 5'=  1330  7' 53/ 

i3i°46'53".C"=50  3i'7.' 

solution. 

1. 0916,        ^'=0.36276, 

39037'i6",^'=i4o°2  2'44", 

1 1 7°  50' 44",/?'=  1 7°  5' J 6". 


§  50  (page  48). 

(I.)  a  =0.097 1, 
^  =  90°  35'  36", 

(7  =  48'"  9'  24". 
5'  =  o.oo5326i. 


134 


ANSli'f^^S    TO  EXERCISES 


(2.)     f— 14.21 1, 

A  =  76°  20'  5", 

5  =  44°  52' 55". 
5  =  80.962. 

(3.)    ^  =  85.892. 

A  =67°  21' 42", 
C=62°48'  18", 
6'=  3962.8. 

(4.)    a  =0.6767, 
5=  15°  9' 21", 
C=  131'' 19' 39". 
5=0.08141. 

(5.)    f  =  72.87, 

-(4  =  40°  50'  32", 
.fi=ii°  2' 28", 
5  =  422.65. 

§  51  (page  49). 


(I.)  A  = 

55° 

20' 

42". 

B  = 

106 

°  35'  36". 

C= 

18" 

3'  42". 

S  = 

267 

.92. 

i2.)A  = 

34° 

24' 

26", 

£  = 

73° 

14' 

56". 

C= 

:72° 

20' 

36", 

S= 

3.61 

43. 

i3-)A  = 

52° 

20' 

24". 

B  = 

107 

°IS 

»'  14", 

C= 

20° 

20' 

24". 

S  = 

1437- 5- 

(4.)  A  = 

97° 

48' 

, 

B  = 

18° 

21' 

48". 

C  = 

63° 

50' 

12", 

S  = 

193 

•13- 

(5.)  A  = 

54° 

20' 

16", 

B^ 

70^ 

27' 

46". 

C  = 

54^^ 

72' 

, 

S~ 

6090. 

(6.)  A--^ 

35' 

59' 

30", 

5=48^44'  32", 

C=95°i5'56", 
5=0.60709. 

§  52  (page  50), 

(1.)  1 1 16.6  ft. 
(2.)  3081.8  yards. 
(3.)  638.34  ft.. 

14653  sq.  ft. 
(4.)  4.1  and  8.1. 
(5.)  13.27  miles. 
(6.)  6667  ft.    One  solution. 
(7.)  121.97. 
(8.)  44°  2'  56". 
(9.)  32.151  sq.  miles. 
(II.)  54°  29'  12". 
(12.)  a  =  12296  ft., 
<r=  13055  ft. 
(13.)  294.77  ft. 
(14.)  222.1  ft. 
(16.)  42 1 1.8  ft. 
(17.)  72.613  miles. 
(18.)  51.035  ft. 
(19.)  0.85872  miles. 
(20.)  2.98  miles. 
(21.)  1331.2  ft 
(22.)  8.2  miles. 
(23.)  187.39  ft. 
(24.)  0.601 1. 
"^25.)  4.81 12  miles. 
(26.)  60°  51'  8". 
(27.)  37.365  ft. 
(28.)  3.2103  miles. 
(29.)   10.532  miles. 
(30.)  851.22  yards. 
(3'.)  9-5722  miles. 
(32.)  6.1271  miles. 
(33.)  280.47  ft. 

(34.)     !  126    1    M. 


ANSIVERS    TO  EXERCISES 


I3S 


(35-)  4-8i34  miles. 
(36.)  2728.25  ft. 

§  53  (page  56). 

(I.)  30°  =  0.5236, 

450  =  0.7854. 

60°  =  1 .0472, 
1 20°  =  2.0944, 
135°=  2.3562, 
720°=  12.5664, 
990°=  17.2788. 

(2.)  I  =  22°  30', 

^=18° 
10 

i  =  28°  38'  53", 
I  =  100°  16'  4". 

(3.)  1-35.  0.54. 

§  74  (page  73). 

(I.)  sin  4.r  =  4  cos'-r  sin.r 

—  4  cos.r  sin^x, 
cos  4-r  =  cos*  X 

—  6  cos^'x  sin''^  jr-|-sin*-r. 

(2.)  sin  6.ar  =  6  cos*  jr  sin  ;i- 

— 20  cos^.r  sin*jr 
+  6  cos-T  sin*jr, 
cos  6x  =  cos*  X 

—  15  cos*.r  sin'';r 
+  15  cos^x  sin*  X  —  sin*  x. 

a/1 
(3.)  x^=i,    x^  =  l-\-z^, 

•^2  =  -*  +  ^'^.    -^3  = -I. 
^=_i_/V^ 


x^  =  \  —  z 


..V~3 


(4.)  x^=i,  .*-,  =  o.309o-4-zo.95ii, 

JTj  =  —  0.8090  +  Z  0. 5878, 

x^  =  —  0.8090  —  /  0.5878, 
x^  =  o.  3090  —  /  0.95 1 1 . 


(23- 

(24. 

(25. 

(26, 
(27. 
(28. 
(29. 

(30- 
(31- 
(32- 
(33- 
(34. 
(35- 
(36. 
(37- 
(38. 

(39- 

(40. 

(41- 


§  77  (page  78). 

X  =  30°. 
_y  =  30°. 
jf  =  0°  or  45°. 
jr  =  6o°. 

J  =  45°. 


^■  =  45°. 
X  =  30°. 

;i:  =  6o° 

X  =  30°. 

No  angle  <  90°. 

X  =  30°. 

sin  92°  =  cos  2° 

cos  127°  =  —  sin  37° 

tan  320°  =  —  tan  40°, 

cot  350°  =  — cot  10°. 

sin  265"  =  — cos  5° 

tan  171°=—  tan  9°. 

cos.r  =  — 1^33, 

tan.r  =  — -/gV'33, 

cot-ir  =  — 1-/33, 

sec-r  =  — /3V'33, 

esc  X  =  |. 
(42.)  sinx  =  — i-v/55. 

tan  x=:^\/sS' 

cotx  =  /5  V55, 

secx=— |, 

cscx  =  —  5**-;  -y/sS- 
(43.)  sinx  =  — f*^  -V/T3, 

COSX=:—   5*3   -v/13, 

cotr  =  |,  sec,r  =  —  5^/13, 


136 


ANSIVENS   TO   EXERCISES 


(44.)  sin;ir=-VvV74, 

cos  ;ir=  7^^/74,  _ 

tanjr=  —  f,  secx=fV74, 
cscjr=  —  J-\/74. 

(45.)  Quadrant  II  or  IV. 

(46.)   Quadrant  I  or  II. 

(47.)  Quadrant  III  or  IV. 

(48.)  Quadrant  I  or  II. 

(49.)  x—o°,  120^,  180'^,  240°, 

(50.)  x=3o°,  135°,  150%  315°. 

(51.)  x=oP,  90°,  120°,  180°,  240°, 

270°. 

(57-)  o. 
(58.)  a. 
(59)  2{a-b). 

(60.)  K^'-^')- 

§  78  (page  80). 

(I.)  306.32  ft. 

(2.)  831.06  ft. 

(3.)  53^  28' 14". 

(4.)  49.39  ft. 

(5.)  0.43498  mile. 

(6.)  209.53  ft. 

(7.)  7.3188  ft. 

(8.)  37^^  36'  30". 

(9.)   109.28  ft. 

(10.)  502.46  ft. 

(II.)  6799.8  ft. 

(12.)  219.05  ft. 

(13.)  491.76  ft. 

(14.)  50   32' 44". 

(15.)  49' 44' 38"- 

(16.)  34063  ft. 

(17.)  32.326  ft..  29°  6'  35". 

(18.)   5.6569  miles  an  hour. 

(19.)   56.295  tt. 

(20.)   103. 09  ft. 


(21.)  71' 33'  54"- 

(22.)  858,160  miles. 

(23.)  238,850  miles. 

(24.)  2163.4  miles. 

(25.)  90,824,000  miles. 

(26.)  432.08  ft. 

(27.)  60.191  ft. 

(28.)  0.32149  mile. 

(29.)   193.77  ft.,  or  1632.9  ft. 

§  79  (page  83). 
3.416  ft. 
3.7865  ft. 
20.45  ft- 
36.024  ft. 
8.6058  sq.  ft. 
181.23  ^^• 
2.9943  ft. 
5.1311  in. 
25.92  ft. 

92*^  I'  24". 

I. 2491. 

33°  12'  4". 


(I-) 
(2.) 

(3) 
(4-) 

(5-) 
(6.) 

(7-) 
(8.) 

(9-) 

(10.) 

(II.) 

(12.) 

(130 

(14) 

(I5-) 
(16.) 

(I7-) 
(18.) 

(19) 
(20.) 
(21.) 
(22.) 
(23) 

(78.) 
(79) 


1 1 248  ft. 
0.60965  mile. 
1 .3764. 

I  •9755- 
19.882. 
0.9397. 
6.4984. 
3.4641. 
6.1981. 
69.978. 
15.25. 

§  80  (page  84). 

^=90*^,  120"^,  240°,  270°. 
x—o\   20°,  45%  9o'^  loo"^, 
135',  140°,  180%  220", 

225',   260',  2JO'\       315", 
340  . 


ANSWERS    TO  EXERCISES 


t37 


(80.)  x=o°,  30°   90°   I 

ScP,  1800 

(24.)  55.74  ft. 

270°. 

(25.)  247.52  ft. 

(81O  x  =  o°,  45°  120°  '. 

:4o°  225° 

(26.)  556.34  ft. 

270°. 

(27.)  455.12  ft. 

,;82.)  ;r  =  o°,90°  180°,  270°. 

(28.)  18.825  ft. 

(83.)  x  =  o°,  90°,  210°,  330°. 

(29.)  2639.4  ft. 

(84.)  X  =  240°,  300°. 

(30O  396.54  ft. 

(85.)  x  =  2io°,  330". 

(31.)  287.75  ft. 

(86.)  X  =  0°,  90°. 

(32.)  2280.6  ft. 

(87.)  x  =  o°,  180° 

(33.)  64.62  ft. 

(88.)  x  =  oo.  1800. 

(34.)  127.98  ft. 

(89.)  x  =  o°,  90°   120°,  I 

80°  240", 

(350  45-183  ft. 

270°. 

(36.)  4365-2  ft. 

(90O  .r  =  450,  1350,  2250 

315°- 

(37.)  140.17  ft. 

(91,)  x  =  30°,  150°  210° 

330°. 

(38.)  610.45  ft. 

§  81  (page  88) 
(I.)  2145.1  ft. 
(2.)  12.458  miles. 
(3.)  1. 1033  miles. 

• 

(39.)  156.66  ft./ 

(40.)  41°  48'  39"  and  125°  25' 

(41.)  51,288,000. 

(42O  364183- 

(430  1 1 586. 

5/ 

(4.)  1508.4  ft. 
(50  1719-3  yards. 
(60  1.2564  miles. 

(70  1346.3  ft. 
(8.)  387.1  yards. 
(9O  5.1083  miles. 

(44.)  947460. 
(45.)  0.89782. 
(46.)  9929.3. 
(470  751-62  sq.  ft 
(48.)  3145.9. 
(49.)  855.1. 

(loO  3791-8  ft. 

(iiO  44152  ft. 

(12.)  28°  57'  20". 

(50.)  876.34. 

§  88  (page  98). 

(13.)  115.27. 

(I.)  c=  54°  59'  47". 

(14.)  44.358  ft. 

^  =  45°  41' 28", 

(15.)  92.258  ft. 

C=65°45'58". 

(16.)  101°  32'  16". 

(20  C=7,°36'47". 

(17.)  0.83732  mile. 

^  =  95°  22', 

(18O  539-1  ft. 

£•=  71=  32'  14", 

(19O  1.239. 

(3.)  C=64°  14' 30", 

(20.)  152.31  and  238.3. 

C'=ii5°45'3o". 

(21.)  67.110  ft. 

«  =  48=  22'  55", 

(22.)  32.071  ft. 

«'=i3i'  37'  5". 

(23.)  137.78  ft 

c=42'  19'  17". 

138 


ANSWERS   TO  EXERCISES 


^ 

=  137 

'*4o'43". 

(4-) 

C 

=  65' 

49' 

54". 

a 

=  63° 

10 

6". 

b 

=  38° 

59' 

12". 

(5.) 

a 

=  75° 

13' 

I". 

B 

=  58° 

25' 

46". 

C 

=  67° 

27' 

I". 

(6.) 

a 

=  76° 

30' 

37". 

b 

=  65° 

28' 

58," 

c 

=  55° 

47' 

44". 

(7.)^ 

=  54° 

44' 

23". 

b. 

=  64° 

36' 

39". 

c 

=  47° 

57' 

45". 

(8.) 

B 

=  96° 

13' 

23". 

a 

=  73° 

17' 

29". 

C: 

=  70° 

8'  38". 

(9.)^: 

=66° 

58' 

a: 

=  11° 

35' 

49". 

C: 

=  4°  35' 26". 

(10.) 

a-. 

=  61° 

4'  55". 

b: 

=  40° 

30' 

22", 

a 

=  50° 

30' 

32". 

§  99  (page  107). 

(I.)  f  =155°  35'  22", 
B=io°  19'  34", 
(7=  171"  48'  22". 

(2.)  a  =  131°  36' 36". 
^=116°  36' 58", 
c  =  2g°  II'  42", 

(3.)  «  =  107"  7' 45". 
5  =  48°  57' 29". 
C  =  62°  31' 40". 

(4.)  ^  =  62°  54' 43". 
ar=  114°  30'  26", 
t  =  56°39'  10". 


(5.)  ^  =  130°  35' 56". 

^  =  30°  25' 34". 

C  =  3i°26'32". 

(6.)  a  =  98°2i'22", 

^=109°  50' 8", 

f  =  115°  13' 4". 

(7.)  -ff  =  32*»  26' 9", 

a  =  84°i4'32". 

t:  =  5i°6'i2". 

(8.)  a  =  80''  5'  8", 

^  =  70°  10' 36", 

<r  =  i45°5'2". 

(9.)  ^=70°  39' 4". 

5  =  48°  36'  2", 

C=ii9°  15' 2". 

(10.)  a = 40°  0'  12". 

Bz=42°  15'  II", 

C=i2i°36'  19". 

§  100  (page  log). 

(I.)  80.895  sq.  in. 

(2.)  26.869  sq.  in. 

(3,)  158.41  sq.  in. 

(4.)  39533  sq.  miles. 

§  101  (page  112). 

(I.)  5C  =  48°  2'  43", 
AC=  52°  53'  9". 
(2.)  7 :  24  A.M. 

(3.)  4  P.M. 

§  102  (page  114). 

(I.)  30291^  miles. 
(2.)  2229.8  miles. 
(3.)  2748.5  miles. 
(4.)  7516.3  miles. 
(5.)  5109  miles. 


THE  END 


ADVERTISEMENTS 


ELEMENTS    OF    GEOMETRY 

By  ANDREW   W.    PHILLIPS,    Ph.D.,   and    IRVING 
FISHER,  Ph.D.,  Professors  in  Yale  University 


Plane  and  Solid  Geometry, 
Plane  Geometry 
Geometry  of  Space  . 


Elements     of     Geometry. 

Abridged  edition . 
Logarithms  of  Numbers 


ALTHOUGH  this  book  meets  all  the  requirements  in 
geometry  for  entrance  to  all  the  colleges  in  the  country, 
it  aims  to  present  more  than  a  mere  minimum  course. 
Among  its  chief  characteristics  are :  Rigor  of  treatment, 
clearness  of  presentation  both  in  the  form  of  statement  and  in 
the  diagrams,  natural  and  symetrical  methods  of  proof,  and 
richness  and  variety  of  original  problems. 
^  The  geometric,  or  space  axioms,  viz.;  the  straight  line 
axiom,  the  parallel  axiom,  and  the  superposition  axiom  are 
separated  from  those  that  relate  to  magnitudes  in  general,  and 
are  emphasized  as  the  foundation  on  which  the  whole 
geometric  superstructure  is  built. 

^[  The  definitions  are  distributed  through  the  book  as  they 
are  needed,  instead  of  being  grouped  in  long  lists  many  pages 
in  advance  of  the  propositions  to  which  they  apply.  An 
alphabetical  index  is  added  for  easy  reference. 
^  The  constructions  in  the  Plane  Geometry  are  also  distrib- 
uted, so  that  the  student  is  taught  how  to  make  a  figure  at 
the  same  time  that  he  is  required  to  use  it  in  demonstration. 
^  In  the  Geometry  of  Space  the  figures  consist  of  half-tone 
engravings  from  the  photographs  of  actual  models  constructed 
for  use  in  the  class  rooms  of  Yale  University.  By  the  side  of 
these  models  are  skeleton  diagrams  for  the  student  to  copy. 
^  Attention  is  also  called  to  the  theorems  of  proportion,  the 
theory  of  limits,  the  use  of  corollaries  as  exercises  to  supply 
the  need  of  inventional  geometry,  and  the  introduction  to 
modern  geometry. 


AMERICAN     BOOK    COMPANY 
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TEXT-BOOKS   ON   ALGEBRA 

By   WILLIAM    J.    MILNE,    Ph.D.,   LL.D.,    President 
New  York  State  Normal  College,  Albany,   N.  Y. 


ACADEMIC  ALGEBRA  . 


MORE  extended  and  more  comprehensive  than  Milne's 
High  School  Algebra,  this  work  not  merely  states  the 
principles  and  laws  of  algebra,  but  establishes  them 
by  rigorous  proofs.  The  student  first  makes  proper  infer- 
ences, then  expresses  the  inferences  briefly  and  accurately, 
and  finally  proves  their  truth  by  deductive  reasoning.  The 
definitions  are  very  complete,  and  special  applications  and 
devices  have  been  added.  The  examples  are  numerous  and 
well  graded,  and  the  explanations  which  accompany  the 
processes,  giving  a  more  intelligent  insight  into  the  various 
steps,  constitute  a  valuable  feature.  The  book  meets  the 
requirements  in  algebra  for  admission  to  all  of  the  colleges. 


ADVANCED  ALGEBRA. 


THIS  book  covers  fully  all  college  and  scientific  school 
entrance  requirements  in  advanced  algebra.  While 
the  earlier  pages  are  identical  with  the  author's 
Academic  Algebra,  more  than  i6o  pages  of  new  matter 
have  been  added.  Among  the  new  subjects  considered 
are  ;  incommensurable  numbers,  mathematical  induction, 
probability,  simple  continued  fractions,  the  theory  of  num- 
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logarithmic  series,  summation  of  series,  and  the  theory  of 
equations,  including  graphical  representation  of  functions  of 
one  variable,  and  approximation  to  incommensurable  roots. 
Over  5,000  unsolved  exercises  and  problems  are  included  in 
the  book.      The  treatment  is  full,  rigorous,  and  scientific. 


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AN     ELEMENTARY     TEXT- 
BOOK    OF     THEORETICAL 
MECHANICS 

By  GEORGE  A.  MERRILL,  B.S.,  Principal  of  the 
California  School  of  Mechanical  Arts,  and  Director  of 
the  Wilmerding  School  of  Industrial  Arts,  San  Francisco 


MERRILL'S  MECHANICS  is  intended  for  the  upper 
classes  in  secondary  schools,  and  for  the  two  lower 
classes  in  college.  Only  a  knowledge  of  elementary 
algebra,  plane  geometry,  and  plane  trigonometry  is  required 
for  a  thorough  comprehension  of  the  work. 
^  By  presenting  only  the  most  important  principles  and 
methods,  the  book  overcomes  many  of  the  difficulties  now 
encountered  by  students  in  collegiate  courses  who  take  up 
the  study  of  analytic  mechanics,  without  previously  having 
covered  it  in  a  more  elementary  form.  It  treats  the  subject 
without  the  use  of  the  calculus,  and  consequently  does  not 
bewilder  the  beginner  with  much  algebraic  matter,  which 
obscures  the  chief  principles. 

^  The  book  is  written  from  the  standpoint  of  the  student 
in  the  manner  that  experience  has  proved  to  be  the  one 
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to  abide  by  the  fundamental  precepts  of  teaching,  no  one 
method  of  presentation  has  been  used  to  the  exclusion  of 
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^  The  explanation  of  each  topic  is  followed  by  a  few  well- 
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place  tables  of  the  natural  trigonometric  functions  are  included. 


AMERICAN     BOOK     COMl'ANY 

C73) 


ELEMENTS    OF   GEOLOGY 

By  ELIOT  BLACKWELDER,  Associate  Professor  of 
Geology,  University  of  Wisconsin,  and  HARLAN  H. 
BARROWS,  Associate  Professor  of  General  Geology 
and  Geography,  University  of  Chicago. 


AN  introductory  course  in  geology,  complete  enough  for 
college  classes,  yet  simple  enough  for  high  school  pu- 
pils. The  text  is  explanatory,  seldom  merely  des- 
criptive, and  the  student  gains  a  knowledge  not  only  of  the 
salient  facts  in  the  history  of  the  earth,  but  also  of  the  methods 
by  which  those  facts  have  been  determined.  The  style  is 
simple  and  direct.  Few  technical  terms  are  used.  The  book 
is  exceedingly  teachable. 

^[  The  volume  is  divided  into  two  parts,  physical  geology 
and  historical  geology.  It  differs  more  or  less  from  its  prede- 
cessors in  the  emphasis  on  different  topics  and  in  the  arrange- 
ment of  its  material.  Factors  of  minor  importance  in  the  de- 
velopment of  the  earth,  such  as  earthquakes,  volcanoes,  and 
geysers,  are  treated  much  more  briefly  than  is  customary. 
This  has  given  space  for  the  extended  discussion  of  matters 
of  greater  significance.  For  the  first  time  an  adequate  discus- 
sion of  the  leading  modern  conceptions  concerning  the  origin 
and  early  development  of  the  earth  is  presented  in  an  ele- 
mentary textbook. 

^1  The  illustrations  and  maps,  which  are  unusually  numerous, 
really  illustrate  the  text  and  are  referred  to  definitely  in  the 
discussion.  They  are  admirably  adapted  to  serve  as  the  basis 
for  classroom  discussion  and  quizzes,  and  as  such  constitute  one 
of  the  most  important  features  of  the  book.  The  questions  at 
the  end  of  the  chapters  are  distinctive  in  that  the  answers  are 
in  general  not  to  be  found  in  the  text.  They  may,  how-, 
ever,  be  reasoned  out  by  the  student,  provided  he  has  read 
the  text  with  understanding. 


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By  D.  D.  MAYNE,  Principal  of  School  of  Agriculture  and 
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THIS  course  has  a  double  value  for  pupils  in  the  first 
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most  profitably.  On  the  other  hand,  it  affords  an  interesting 
introduction  to  all  the  natural  sciences,  enabling  the  student  to 
master  certain  definite  principles  of  chemistry,  botany,  and 
zoology,  and  to  understand  their  applications.  A  few  experi- 
ments are  included,  which  may  be  performed  by  the  student 
or  by  the  teacher  before  the  class.  But  the  subject  is  not 
made  ultrascientific,  forcing  the  student  through  the  long 
process  of  laboratory  method  to  rediscover  what  scientists 
have  fully  established. 

^  The  topics  are  taken  up  in  the  text  in  their  logical  order. 
The  treatment  begins  with  an  elementary  agricultural  chem- 
istry, in  which  are  discussed  the  elements  that  are  of  chief 
importance  in  plant  and  animal  life.  Following  in  turn  are 
sections  on  soils  and  fertilizers;  agricultural  botany;  economic 
plants,  including  field  and  forage  crops,  fruits  and  vege- 
tables ;  plant  diseases ;  insect  enemies ;  animal  husbandry ;  and 
farm  management. 

^[  The  chapter  on  plant  diseases,  by  Dr.  E.  M.  Freeman, 
Professor  of  Botany  and  Vegetable  Pathology,  College  of 
Agriculture,  University  of  Minnesota,  describes  the  various 
fungus  growths  that  injure  crops,  and  suggests  methods  of 
fighting  them.  The  section  on  farm  management  treats  farm- 
ing from  the   modern  standpoint  as  a    business  proposition. 


AMERICAN    BOOK    COMPANY 

(3M) 


A    BRIEF    COURSE    IN 
GENERAL    PHYSICS 


By    GEORGE    A.     HOADLEY,    A.M.,    C.E. 
Professor  of  Physics,  Swarthmore  College 


A  COURSE,  containing  a  reasonable  amount  of  work  for 
an  academic  year,  and  covering  the  entrance  require- 
ments of  all  of  the  colleges.  It  is  made  up  of  a  reliable 
text,  class  demonstratibns  of  stated  laws,  practical  questions 
and  problems  on  the  application  of  these  laws,  and  laboratory 
experiments  to  be  performed  by  the  students. 
^  The  text,  which  is  accurate  and  systematically  arranged, 
presents  the  essential  facts  and  phenomena  of  physics  clearly 
and  concisely.  While  no  division  receives  undue  prominence, 
stress  is  laid  on  the  mechanical  principles  which  underlie  the 
whole,  the  curve,  electrical  measurements,  induced  currents, 
the  dynamo,  and  commercial  applications  of  electricity. 
^  The  illustrative  experiments  and  laboratory  work,  intro- 
duced at  intervals  throughout  the  text,  are  unusually  numerous, 
and  can  be  performed  with  comparatively  simple  apparatus. 
Additional  laboratory  work  is  included  in  the  appendix,  to- 
gether with  formulas  and  tables. 


HOADLEY'S    PRACTICAL    MEASUREMENTS    IN 
MAGNETISM  AND  ELECTRICITY. 

THIS  book,  which  treats  of  the  fundamental  measurements  in  elec- 
tricity as  applied  to  the  requirements  of  modern  life,  furnishes  a  satis- 
factory introduction  to  a  course  in  electrical  engineering  for  secondary 
and  manual  training  schools,  as  well  as  for  colleges.  Nearly  loo  experiments 
are  provided,  accompanied  by  suggestive  directions.  Each  experiment  is 
followed  by  a  simple  discussion  of  the  principles  involved,  and,  in  some 
cases,  by  a  statement  of  well-known  results. 


AMERICAN     BOOK     COMPANY 


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ELEMENTS    OF 
POLITICAL    ECONOMY 

By  J.  LAURENCE    LAUGHLIN,    Ph.D.,    Head    Pro- 
fessor ot  Political  Economy,  University  of  Chicago 


IN  the  present  edition  the  entire  work  is  thoroughly  revised 
and  as  regards  both  theory  and  practical  data  is  entirely 
in  accord  with  the  times.       The  treatment  is  sufficiently 
plain  for  even  high  school  students. 

^  The  book  is  in  two  parts:  Part  I,  pertaining  to  the  prin- 
ciples of  political  economy  and  containing  chapters  on  the 
many  phases  of  production,  exchange,  and  distribution;  and 
Part  II,  treating  of  such  important  topics  as  socialism,  taxation, 
the  national  debt,  free  trade  and  protection,  bimetallism.  United 
States  notes,  banking,  the  national  banking  system,  the  labor 
problem,  and  cooperation. 

^  The  work  is  equally  suitable  for  a  short  or  a  long 
course,  and  contains  many  valuable  practical  exercises  which 
are  intended  to  stimulate  thought  on  the  part  of  the  stu- 
dent, A  large  bibliography,  footnotes,  and  references  are 
included. 

^  Throughout  the  main  purpose  is  to  present  a  fair  and 
impartial  discussion  of  the  important  questions  of  the  day,  and 
to  give  a  large  amount  of  useful,  practical  information,  rather 
than  to  devote  extended  space  to  abstract  theory. 
^  Among  the  important  features  of  the  new  edition  are  a 
discussion  of  the  law  of  satiety,  final  utility,  and  its  relation- 
ship to  expenses  of  production  in  the  theory  of  value;  an 
explanation  of  the  industrial  system  wherein  the  time  element 
has  created  a  different  organization  from  that  of  primitive 
society;  an  adjustment  of  consumption  to  the  general  eco- 
nomic principles;  an  enlarged  statement  of  the  development 
of  division  of  labor,  and  a  brief  discussion  of  large  production 
and  so-called  "trusts." 


AMERICAN     BOOK    COMPANY 

(189) 


CHEMISTRIES 

By  F.  W.  CLARKE,  Chief  Chemist  of  the  United  States 
Geological  Survey,  and  L.  M.  DENNIS,  Professor  of 
Inorganic  and  Analytical  Chemistry,  Cornell  University 


Elementary   Chemistry 


Laboratory  Manual 


THESE  two  books  are  designed  to  form  a  course  in 
chemistry  which  is  sufficient  for  the  needs  of  secondary 
schools.  The  TEXT-BOOK  is  divided  into  two  parts, 
devoted  respectively  to  inorganic  and  organic  chemistry. 
Diagrams  and  figures  are  scattered  at  intervals  throughout  the 
text  in  illustration  and  explanation  of  some  particular  experi- 
ment or  principle.  The  appendix  contains  tables  of  metric 
measures  with  English  equivalents. 

^  Theory  and  practice,  thought  and  application,  are  logically 
kept  together,  and  each  generalization  is  made  to  follow  the 
evidence  upon  which  it  rests.  The  application  of  the  science 
to  human  affairs,  its  utility  in  modern  life,  is  also  given  its 
proper  place.  A  reasonable  number  of  experiments  are  in- 
cluded for  the  use  of  teachers  by  whom  an  organized  laboratory 
is  unobtainable.  Nearly  all  of  these  experiments  are  of  the 
simplest  character,  and  can  be  performed  with  home-made 
apparatus. 

4  The  LABORATORY  MANUAL  contains  127  experi- 
ments, among  which  are  a  few  of  a  quantitative  character.  Full 
consideration  has  been  given  to  the  entrance  requirements  of 
the  various  colleges.  The  left  hand  nages  contain  the  experi- 
ments, while  the  right  hand  pages  are  left  blank,  to  include 
the  notes  taken  by  the  student  in  his  work.  In  order  to  aid 
and  stimulate  the  development  of  the  pupil's  powers  of  observa- 
tion, questions  have  been  introduced  under  each  experiment. 
The  directions  for  making  and  handling  the  apparatus,  and 
for  performing  the  experiments,  are  simple  and  clear,  and  are 
illustrated  by  diagrams  accurately  drawn  to  scale. 


AMERICAN     BOOK     COMPANY 


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